1. Why Treat Wastewater?

1.1 Deﬁnition of Wastewater

Wastewater is human-polluted water from home and industries. This includes water from:

• Flushing toilets and urinals - blackwater.

• Bathing, showering, and washing clothes and dishes - greywater.

• Commercial and industrial activities.

•

...and often included is stormwater which contain pollutants washed off from inhabited

areas - roads, parking lots, and rooftops.

1.2 Why Treat Wastewater

Although nature has an inherent capability to breakdown pollutants, given the large quantity gen-

erated from human activities, there is a need for centralized wastewater treatment plants to treat

wastewater before releasing it back to the environment. Wastewater from homes, businesses, and

industries are collected in sewers for delivery to the treatment plant and subsequently discharged

to a water body like a lake, river or ocean, or land, or reused.

Wastewater treatment is designed to remove:

• organic matter

• inorganic pollutants including plant nutrients - nitrogen and phosphorous

• pathogenic (disease causing) organisms

1.3 Beneﬁts of Treating Wastewater

Wastewater treatment protects:

• The environment

Chapter 1. Why Treat Wastewater?

• Human health

Speciﬁcally, wastewater treatment allows for the following:

1. Mitigates deterioration of the receiving waters’ ecosystem

The discharge of inadequately treated wastewater will cause oxygen levels in the receiving

waters to be depleted, due to:

•

Wastewater containing nitrogen and phosphorus based pollutants (plant nutrients)

entering a water body such as a lake or river will promote plant and algae growth

which will seriously impact its normal aquatic life including ﬁsh through a process

similar to the following:

– Nutrient promote algae bloom

–

Algae bloom prevent sunlight to the native plant spieces below the water’s sur-

face causing native plants to die

–

The organic material from the dead plants and algae promote growth of aerobic

bacteria which will consume the dissolved oxygen in the water resulting in

oxygen depletion.

–

The natural aquatic life including ﬁsh, frogs, and turtles will not be able to sur-

vive under oxygen depleted conditions and will die or leave that zone.

•

Other organic material present in wastewater will similarly promote growth of aero-

bic bacteria, intensifying the eutrophication of the receiving waters.

Thus, proper treatment of wastewater will prevent eutrophication - which is the depletion

of dissolved oxygen of the receiving water, consequently impacting/destroying its normal

aquatic life.

2. Removal of other harmful pollutants

Organic and inorganic pollutants, including metals such as mercury, lead, cadmium,

chromium and arsenic can have acute and chronic toxic effects on aquatic species and

wildlife including migratory birds, are removed during the wastewater treatment process.

3. Removal of pathogens

Wastewater treatment removes parasites and disease-causing pathogens including bacteria

and viruses for:

•

People to continue enjoying recreational activities in the receiving bodies of waters

such as lakes and rivers

•

Preventing the contamination of ﬁsh and other consumable products obtained from

the waters

• Allow the water body to remain as the source of potable water

4. Reclaim water for recycle or reuse

Besides protecting human health and the environment, wastewater treatment paves way

for establishing the reuse or recycle of treated wastewater. This beneﬁt is particularly

important for densely populated areas with limited access to fresh water supplies.

2. Regulations

2.1 Regulations Related to Wastewater Treatment

The main objective of these regulations is to ensure appropriate quality of the treated wastewater.

2.1.1 Treated Wastewater - NPDES Permit

The National Pollutant Discharge Elimination System (NPDES) permit program program ad-

dresses water pollution by regulating point sources that discharge pollutants to waters of the

United States.

•

The NPDES permit program was created in 1972 by the Clean Water Act (CWA)and is

administered by the federal USEPA.

• Applies to sources that discharge pollutants to waters of the United States.

•

Requires all facilities discharging “pollutants” into any body of water in the USA to obtain

and comply with a NPDES permit.

• NPDES permit establishes discharge limits, monitoring and reporting requirements

•

In California, the responsibility of implementing the federal NPDES program is delegated

to the State of California through the State Water Resources Control Board (State Water

Board or SWRCB) and ﬁnally to the nine Regional Water Quality Control Boards (Re-

gional Water Boards or RWQCB), collectively known as Water Boards.

• The RWQCB issues the NPDES permit.

2.1.2 Inﬂuent Wastewater - National Pretreatment Program

Municipal wastewater treatment plants also known as Publicly Owned Treatment Works (POTWs)

implement and enforce their Pretreatment or Industrial Discharge Control programs to meet Fed-

eral and State regulations requirements related to wastewater discharges from industrial sources.

Chapter 2. Regulations

The national pretreatment program is a component of the NPDES program and is also known as

the Source Control Program

The Pretreatment/Source Control program is for controlling industrial (non-domestic) wastewater

discharges with the following objectives:

1. Protect the treatment plant operations so that the industrial discharge does not contain pol-

lutants or have certain characteristics (including pH, temperature) which could adversely

effect the treatment process or impact public safety and the safety of the people working at

the treatment plant.

2. Prevent the introduction of pollutants that could pass through untreated and into the receiv-

ing body of water.

3. Improve opportunities for reuse or recycling of wastewater and sewage sludge.

In California:

1. Wastewater treatment plants are required to have a Pretreatment Program when their total

design ﬂows are greater than ﬁve million gallons per day (5 mgd).

2. Facilities with smaller ﬂows (5 mgd or less) may also be required to implement a Pretreat-

ment Program if they receive industrial waste and pretreatment is warranted.

3. The Pretreatment Program for a wastewater treatment entity is reviewed and approved by

the State and Regional Water Boards, and

4. The Pretreatment Program’s monitoring and reporting requirements are incorporated in the

facility’s NPDES permit.

2.2 Sewage Sludge/Biosolids Regulations

Sewage sludge or biosolids is a byproduct of wastewater treatment. The biosolids produced

are disposed or used using methods including land application, landﬁll and incineration. Fed-

eral Regulation 40CFR Part 503 also known as Rule 503 establishes the standards for the use

or disposal of wastewater biosolids - as stipulated under the Clean Water Act. The facility’s

NPDES permit incorporates the applicable federal, state and local requirements as they apply to

its biosolids.

•

Part 503 rule applies to any person who applies biosolids to the land or ﬁres biosolids in a

biosolids incinerator, and to the owner/operator of a surface disposal site, or to any person

who is a preparer or generator of biosolids for use, incineration, or disposal.

• Part 503 standard includes:

1. General requirements which establishes the purpose and applicability of the rule, the

compliance period, and exclusions from the rule.

2. Limits on heavy metals content

3. Solids management practices related to use and disposal of wastewater biosolids

4. Operational standards related to biosolids management, and

5. Requirements for the frequency of monitoring, record-keeping, and reporting

Part 503 requirements are factored in when establishing the heavy metals concentration limits for

2.3 Air Quality Regulations

the Pretreatment or Industrial Control Program as a signiﬁcant portion of the heavy metals in the

inﬂuent wastewater are removed as part of the wastewater solids.

2.3 Air Quality Regulations

•

Air emissions from wastewater collections and treatment systems are subject to federal,

state and local air quality related rules and regulations established to protect human health

and comfort, and the environment.

•

Typically, a local agency such as the South Coast Air Quality Management District (SCAQMD)

is designated to enact and enforce air quality rules and regulations, through its permitting

process, applicable to all sources of air emissions including wastewater treatment plants.

• Systems/processes subject to air quality regulations at air quality regulations include:

–

Fugitive emissions: Foul air containing compounds such as hydrogen sulﬁde and

organics, which escape from process tanks, pipes and associated structures such as

manholes and wetwells, is potentially harmful for the affected public and also cause

odors.

–

Digester gas combustion: Digester gas a product of wastewater solids treatment

contains methane and is either combusted in power generation equipment or burned

in ﬂares.

Odor control systems: These commonly used systems are for controlling emissions

of regulated pollutants such as ammonia and to prevent odors associated with com-

pounds such as hydrogen sulﬁde.

• Related to its air pollutants emissions, wastewater treatment plants are required to:

–

Obtain air quality related operating permits for equipment and processes which emit

air pollutants and for its systems treating foul air.

– Implement air emission pollutants control measures

– Comply with record keeping and reporting requirements

–

Comply with air quality rules to prevent public nuisance and protect public health

and safety

2.4 Regulations Related to Operations and Maintenance

2.4.1 Operator Certiﬁcation

•

The requirements of the Operator Certiﬁcation program is established for each state. These

meet the Operator Certiﬁcation Requirements of the regulations stemming from the 1996

Amendments to the Safe Drinking Water Act.

•

The goal is to ensure that operators of wastewater treatment facilities in the State meet the

minimum level of competence; thereby, protecting public health and the environment.

In California, the Wastewater Operator Certiﬁcation program (WWOCP) administers

Wastewater Treatment Plant Certiﬁcation examinations, certiﬁcations (grades I to V), and

certiﬁcation renewals.

Chapter 2. Regulations

•

WWOCP classiﬁes Wastewater Treatment Plants and stipulates that no person shall oper-

ate a wastewater treatment plant unless that person has been certiﬁed by the division as a

wastewater treatment plant operator or operator-in-training at a grade appropriate for the

class of plant being operated.

A certiﬁed operator or operator-in-training may be subject to administrative sanctions

including reprimand or denial, suspension, probation, or revocation of the operator certiﬁ-

cation for performing, or allowing or causing another to perform acts which include:

–

Operating or allowing the operation of a wastewater treatment plant by a person who

is not certiﬁed at the grade necessary for the position

failing to use care or good judgment in the course of employment as an operator or

failing to apply knowledge or ability in the performance of duties.

Negligence causing the violation of appropriate waste discharge requirements of the

NPDES permit

Classification of Wastewater Treatment Plants

Grade Levels of Operator Certifications

Chapter 2. Regulations

2.4.2 Worker Safety

• Wastewater treatment facility can be an extremely unsafe occupational ﬁeld

•

It involves most of the major categories of workplace hazards: biological, chemical, phys-

ical, safety and ergonomic, accentuated with other factors such as shift work and diverse

tasks.

•

Entities including The Occupational Safety and Health Administration(OSHA) National

Electrical Code (NEC), National Fire Protection Association (NFPA), Underwriters Labo-

ratory (UL) have recognized these hazards and implemented codes and standards to protect

the affected persons and wastewater workers.

3. Elements of Wastewater Treatment

Wastewater cycle is a part of the water cycle where the water consumed as part of the normal

human and industrial activity is returned back to the environment after treatment.

Wastewater cycle comprises of the following sequential elements:

1. Generation

2. Collection

3. Treatment

4. Disposal or reuse

3.1 Generation

Wastewater originates from domestic, industrial, commercial or agricultural activities. The char-

acteristics of wastewater vary depending on the source. Types of wastewater include:

•

Domestic Sewage: wastewater derived principally from dwellings, business buildings,

institutions, and

• Industrial Sewage: liquid waste from industrial processes

Typical per person generation of wastewater in the USA is about 70-100 gallons per day

3.2 Collections

•

Wastewater is collected from its point of origin - home, businesses, industries etc. and

conveyed via sewer lines to a centralized wastewater treatment facility.

When the rainwater drainage is made part of the sewer system, the system is termed as

Combined System.

•

Chapter 3. Elements of Wastewater Treatment

•

The system where the sewage is conveyed separately from the stormwater ﬂows is termed

as Separated System.

In the Separated System, the Sanitary Sewers convey the wastewater and the Stormwater

Sewer conveys the storm water ﬂows.

For the Combined System, rainstorms pose the threat of overwhelming the sewers and the

treatment plant

Combined System

Separated System

3.3 Treatment

3.3.1 Liquid Phase Treatment

•

Wastewater treatment can involve physical, chemical or biological processes or combina-

tions of these processes depending on the required outﬂow standards.

• Wastewater treatment typically involves a series of steps with increasing level of treatment:

–

Preliminary: The preliminary process removes large/coarse solids which include

rocks, tree branches, grit and other debris present in wastewater.

Primary: The primary process is also a physical process where the separable wastew-

ater solids - solids that ﬂoat and solids that can settle, are removed.

Secondary: Secondary treatment is a biological treatment process where microorgan-

isms consume the organic matter present in the wastewater.

Tertiary or Advanced Treatment: The tertiary/advanced treatment processes improve

the quality of treated water beyond the secondary treatment level. This process may

include nutrient removal and disinfection.

3.3.2 Treatment of Wastewater Solids

– Solids are a byproduct of wastewater treatment.

–

Screenings and grit removed as part of the preliminary treatment is typically disposed

in a landﬁll.

–

Sludge generated from the wastewater treatment processes - settled solids and scum

from primary and secondary treatment processes needs to be treated prior to disposal

or reuse to comply with wastewater solids - biosolids regulations.

3.3 Treatment

Typical solids treatment is comprised of the following three sequential steps:

1. Sludge thickening

2. Sludge stabilization

3. Sludge dewatering

Sludge Thickening

Sludge thickening improves performance of sludge stabilization process and provides

capital and operational cost savings due to a lower volume of sludge

Sludge Stabilization

Sludge stabilization process produces solids (biosolids) that meet Part 503 rule require-

ments.

Sludge Dewatering

Solids stabilized using digestion process has only a small percentage by weight of solids

-less than 5%. It therefore becomes necessary to dewater the stabilized sludge prior to

hauling off-site for ﬁnal disposal. A generalized layout/process sequencing in a wastewater

treatment plant is shown below:

Individual wastewater treatment processes involve different process options or sequences

which are illustrated in the graphic below:

Chapter 3. Elements of Wastewater Treatment

3.4 Disposal or Reuse

•

Wastewater treatment processes can be designed to dispose the treated water where the

water is reintroduced to the environment or for reuse where the treated water is reclaimed

or recycled - for various purposes including irrigation, industrial use or for potable use.

• Water disposal methods include:

– Surface water discharge

– Subsurface discharge

• Water reuse methods include:

– Potable water reuse

Indirect potable reuse: Here the treated water is blended with groundwater or

surface water and then reclaimed and treated further for drinking (potable) water

use

Direct potable reuse: Here the treated wastewater is subjected to advanced

treatment and introduced directly into a municipal water supply system

– Water reclamation for irrigation or industrial use

– Land application for beneﬁcial use

•

Solids generated from the wastewater treatment process may be removed and disposed to

a landﬁll or subject to further treatment which may allow for energy recovery - from the

organic solids and for beneﬁcial reuse due to its plant nutrient content.

4. Constituents Properties and Analysis

4.1 Wastewater Constituents

4.1.1 Organics

• The main reason for treating domestic wastewater is to remove the organic matter.

•

Organics are substances containing carbon, hydrogen and oxygen, and some of which may

be combined with nitrogen, sulfur or phosphorous.

•

About 50 percent of the solids present in wastewater are organic. This fraction is generally

of animal or vegetable life, dead animal matter, plant tissue or organisms, and also include

synthetic organic compounds.

•

The principal organic compounds present in domestic wastewater are proteins, carbohy-

drates and fats together with the products of their decomposition.

Organics are subject to decay or decomposition through the activity of bacteria and other

living organisms. Since the organic fraction can be driven off at high temperatures, they are

also called volatile solids.

•

Organics in wastewater is typically quantiﬁed in terms of oxygen required to oxidize the

carbon based material present in wastewater using the following methods:

Biochemical Oxygen Demand (BOD)

– Oxygen is required for the consumption of organic matter by aerobic bacteria

–

BOD test measures the depletion of oxygen in a wastewater sample over a ﬁve day

period

BOD measures the organic content in terms of oxygen required for the microorgan-

isms to consume the organic material present

BOD is typically measured as BOD₅which is the oxygen demand of the wastewater

Chapter 4. Constituents Properties and Analysis

measured after 5 days of the initiation of the test.

–

The test involves incubating a known dilution of wastewater in a 300 ml bottle for 5

days at 20^◦C. The dissolved oxygen (DO) content at the start and end of the incuba-

tion period is used for calculating the BOD.

For the test to be considered valid, the following criteria need to be met: 1) DO con-

sumption during the test must be at least 2 mg/l, 2) DO remaining at the end of the

test must be at least 1 mg/l, and 3) DO consumed in blank should be 0.2 mg/l or less

BOD is a parameter to measure the strength of wastewater and the measurement of

the wastewater treatment plant or treatment process inﬂuent and efﬂuent BOD is

standard practice to measure its performance. Typical domestic wastewater BOD is

about 200-250 mg/l.

–

The oxygen consumed by the microorganisms during the BOD test is primarily for:

1) Oxidizing the carbonaceous material (cBOD – carbonaceous BOD), and 2) Oxidiz-

ing nitrogenous constituents such as ammonia (nBOD – nitrogenous BOD).

Thus, BOD (Total) = cBOD + nBOD. The cBOD and nBOD is measured by adding

certain chemical inhibitors which will inhibit the bacteria responsible for consuming

the nitrogenous matter, thus measuring only the cBOD as part of the BOD test.

Since not all of the organics is metabolized in the 5 days of the regular BOD test,

certain wastewater discharge permits require reporting of the ultimate BOD value

(BOD_U)

Chemical Oxygen Demand (COD)

–

The COD test involves using chemical oxidizers to measure the oxygen demand of

the wastewater.

As the chemical oxidizers will oxidize other constituents present, including inorganic

matter, the COD value of wastewater will be higher than the BOD.

The COD test can be conducted rather quickly than the 5 day BOD test, it is an effec-

tive method to quantify the wastewater strength and process efﬁciencies and allow

operators to make timely process adjustments.

Total Organic Carbon (TOC)

The TOC method utilizes laboratory analytical instruments which directly measure the

organic carbon content by quantifying the amount of carbon dioxide produced from the

complete combustion of the organics present.

Note: BOD measures the amount of oxygen required by the microorganisms present to consume

the organic material while COD measures the chemical oxidation required to oxidize all chemicals

including organics present in wastewater. BOD value of typical domestic sewage is about 200 -

250 mg/l while the COD value ranges from 300 - 450 mg/l. Typical BOD:COD ratio ranges from

0.5-0.8.

4.1 Wastewater Constituents

4.1.2 Solids

Like BOD, wastewater solids is another critical parameter for establishing the wastewater strength

and determining treatment process efﬁciencies.

•

The solids can be classiﬁed as suspended or dissolved based upon its ability to pass through

a standardized ﬁlter paper.

• When the wastewater is ﬁltered:

–

the residual solids remaining on the ﬁlter paper after drying in an oven at 103^◦C is

the suspended solids portion, and

– the solids remaining after drying the ﬁltrate are the dissolved solids.

•

Suspended solids include larger ﬂoating particles and consist of sand, grit, clay, fecal

matter, paper, pieces of wood, particles of food and garbage, and similar materials.

• Suspended solids can be categorized based upon its settling characteristics as:

– Settleable

– Non-settleable

Colloidial-small, charged (typically negative) particles which do not settle easily.

Some of the colloidial particles are small enough to pass through the ﬁlter paper

used for ﬁltering the suspended solids

Floatable-example oil and grease and small plastics

•

Dissolved solids in wastewater include organics. However, the major elements of dissolved

solids are inorganic ions such as Ca⁺², Mg⁺², Cl⁻, SO₄⁻², HCO₃⁻, Fe⁺², PO₄⁻³, NO₃

⁻. These ions are part of the dissolved salts such as sodium chloride (NaCl), calcium bi-

carbonate (Ca(HCO₃)₂), magnesium phosphate (Mg₃PO₄) and others which are normally

present in water and wastewater.

–

Conductivity or electrical conductance (EC) measurement is typically conducted as

the wastewater enters the plant as conductivity provides an indirect and simple measure

of the amount of dissolved solids present.

–

Conductivity or electrical conductance (EC) is a measure the amount of electrical

current that can be conducted by a solution.

The conductance of electricity in a solution is due to the presence of dissolved inor-

ganic ions

– The higher the concentration of these ions, the higher is the conductivity.

– Conductivity is measured in the units of mhos/cm or Siemens/cm. (Note: mhos is the

reverse of ohm which is a measure of resistance).

– Typical wastewater conductivities range from 50 to 1500 S/cm

• Both suspended and dissolved solids can be either volatile (organic) or ﬁxed (inorganic).

• Total Solids is thus a sum of TSS and dissolved solids or volatile and ﬁxed solids.

– The volatile solids are typically of plant or animal origin .

– The ﬁxed solids include sand, gravel and silt as well as the dissolved salts.

Chapter 4. Constituents Properties and Analysis

The volatile or ﬁxed fractions are quantiﬁed

•

by incinerating the solids in a mufﬂer furnace

at 550^◦which removes only the volatile solids

leaving only the ﬁxed solids behind.

In terms of the size of the solids, the distribu-

tion is approximately thirty percent suspended

and about seventy percent dissolved solids -

which includes the colloidal particles which

have passed through the ﬁlter paper.

•

As primary treatment process involve settling of

solids, establishing the settleable portion of the

suspended solids is important.

Imhoff Cone

The settleable solids are quantiﬁed using an

Imhoff cone and are reported in ml/L. Imhoff

cone is a 1 liter, clear cone shaped container,

with volume graduations (ml) at the bottom.

One factor which affects settleability is the conveyance time of the sewage to the treatment

plant.

Note the ml markings at the bottom of the cone

•

The settleable component of the suspended solids will decrease as the sewage becomes

more septic due to longer conveyance times.

Inﬂuent and efﬂuent total suspended solids are measured to establish the overall treatment

and individual process efﬁciencies.

Volatile solids measurements before and after biological processes such as secondary

treatment and digestion provide information on the process efﬁciency.

Summary of Wastewater Solids

• Solids in wastewater can be categorized as dissolved or suspended

– Suspended solids can be further categorized as settleable or unsettleable

• Solids can also be categorized as organic (aka: volatile) or inorganic (aka: ﬁxed)

•

Colloidial particles are small sized particles some of which pass through the ﬁlter and

accounted as part of dissolved solids

TSS - Total Suspended Solids are the solids that are captured on the ﬁlter paper upon

ﬁltration of the wastewater sample.

Wastewater samples typically analyzed for TSS include: plant, primary and secondary

processes - inﬂuent and efﬂuent. TSS is reported in mg/l

TS - Total Solids are solids content of sludge. TS of sludge is established by drying a

preweighed quantity of sludge in an oven and is typically reported as % solids - which is

how many parts (by weight) of solids per 100 parts (by weight) of sludge.

Volatile solids are solids that are removed when the solids are incinerated at 550C. The

solids that remain after incineration are ﬁxed or non-volatile or inorganic solids.

•

4.1 Wastewater Constituents

Wastewater Solids Content

•

Less than 0.1% total solids. Total solids concentration in typical wastewater is about

750mg/l

• The total solids are 50% organic (volatile) and 50% inorganic (ﬁxed)

Of the total solids, dissolved solids constitute about 70% of the solids and the remaining

30% solids are suspended solids

•

• 40% of the dissolved solids are volatile the remaining 60% are ﬁxed

• 70% of the suspended solids are volatile and the remaining 30% are ﬁxed

Figure 4.1: Typical Wastewater Solids Concentrations

4.1.3 Nutrients

•

Plant nutrients - nitrogen and phosphorous, present in wastewater efﬂuent discharge, pro-

mote growth of plant and algal matter in the receiving waters causing destruction of the

normal aquatic life mainly due to oxygen depletion - eutrophication.

Because of the potential impacts of the presence of these nutrients in wastewater efﬂuent

on the receiving waters, limits on the levels of these nutrients is typically stipulated in the

treatment plant’s wastewater discharge permit.

Typically, conventional secondary treatment processes are designed primarily remove

the organics from the wastewater. Secondary treatment process designed to additionally

remove nutrients is deemed as tertiary or advanced treatment is termed as Biological Nutri-

ent Removal (BNR).

Chapter 4. Constituents Properties and Analysis

Nitrogen

Forms of nitrogen:

•

About 60% of nitrogen in wastewater is present as ammonia nitrogen (about 60%). The

ammonium nitrogen is present either in the form of ammonia (NH₃) or as ammonium

(NH⁺₄) ion. These two forms can rapidly change from one to the other depending on pH

and temperature. Under low pH (acidic) or neutral conditions – pH less than or equal to 7,

ammonia exists mostly as ammonium. Ammonia becomes the dominant form as the pH

increases to 8 and beyond.

•

The other dominant form of nitrogen, about 40% of the total nitrogen is as organic nitro-

gen

Nitrogen measured as Total Kjeldahl Nitrogen (TKN) which is the sum of the organic

nitrogen and the ammonia nitrogen concentrations. Total inorganic nitrogen is the total

concentration of ammonia nitrogen, NO3-, and NO2-. Table provides the concentrations

and forms of nitrogen in wastewater.

Forms of Nitrogen in Wastewater

Forms of Nitrogen

Formula

Found in

Typical

Concentra-

tion

30-50 mg/l

Ammonia/Ammonium

NH₃/NH₄

Inﬂuent wastewa-

ter

Wastewater

efﬂuent

Total Kjeldahl Nitrogen

(Ammonia/Ammonium + Organic Nitro-

TKN

30-60 mg/l

gen)

Total Inorganic Nitrogen

TIN

Wastewater

efﬂuent

1-40 mg/l

(Ammonia/Ammonium + Nitrite + Ni-

trate)

Nitrate

−

NO₃₋

Nitriﬁed efﬂuent

Partially nitriﬁed

efﬂuent

1-35 mg/l

0.1-2 mg/l

NO₂

Figure 4.2: Forms of Nitrogen

Phosphorous

Forms of phosphorous:

•

The principal forms are organically bound phosphorus, polyphosphates, and orthophos-

phates.

Organically bound phosphorus originates from body and food waste and, upon biological

decomposition of these solids, is converted to orthophosphates.

Polyphosphates originate from synthetic detergents and are hydrolyzed to orthophosphates.

Thus, the principal form of phosphorus in wastewater is assumed to be orthophosphates,

3−

although the other forms may exist. Orthophosphates consist of the negative ions PO₄

4.2 Wastewater Properties and Parameters

HPO₄²⁻, and H₂PO₄⁻. These may form chemical combinations with cations (positively

charged ions).

Oil and Grease

Fats, oil and grease in wastewater originate from homes, food establishments and industries.

•

Oil and grease content of wastewater is established in the laboratory by extracting it with a

solvent - n-hexane. The concentration of oil and grease is reported in mg/l and typical oil

and grease content of wastewater ranges from 80 - 120 mg/l

Presence of excessive oils and grease could potentially impact the secondary treatment

process

•

Oils and grease are removed as ﬂoatables in primary treatment and sent with the sludge to

the digesters

4.2 Wastewater Properties and Parameters

Laboratory and ﬁeld tests are conducted to measure parameters which are critical for monitoring

and controlling treatment. The following are the key parameters that are measured.

4.2.1 pH

pH is a measure of the hydrogen ion (H⁺) content or the acidity or basicity of a solution. pH

impacts the chemical and micribiological elements of wastewater treatment processes and thus

pH measurement and control is critical.

• Pure water dissociates into equal concentration of hydrogen ions and hydroxide ions:

H₂O → H⁺+OH⁻.

• The H⁺are responsible for acidic properties and the OH⁻ions for the basic properties.

•

pH is the inverse of H⁺concentration; pH increases when the concentration of H⁺de-

creases relative to the concentration of OH-.

pH scale ranges from 0 – 14. When the concentration of both H⁺and OH⁻are equal, as in

pure water, it is considered neutral and its pH is 7.0.

If the pH of a sample solution is below 7.0, the sample is termed acidic and is alkaline or

basic if its pH is above 7.0.

Each change of 1 pH unit represents a 10 fold change in concentration. For example, a

sample with a pH of 2.0 is 1000 times more acidic than a sample with a pH of 5.0.

pH is measured by an electrode that is sensitive only to H⁺or using a pH strip which is

essentially an adsorbent paper which is pre-impregnated with chemicals which change

color under different H⁺concentrations.

•

Most organisms involved in biological wastewater treatment processes do well within a a

narrow range of pH near neutral (pH of 7).

Chapter 4. Constituents Properties and Analysis

4.2.2 Oxidation Reduction Potential (ORP)

ORP measurements are common in wastewater process control for monitoring conditions

and process efﬁciency

• ORP is measured in milivolts (mV) using a probe

•

ORP is a measure of the potential of oxidation/reduction – electron transfer, based chemi-

cal reactions to occur

If the measured ORP value (in mV), is positive it indicates an environment where oxida-

tion will occur and if negative, an environment where reduction reactions will occur

Higher positive value indicates a stronger oxidative environment and likewise, a lower

(more negative) ORP value indicates a stronger reductive environment

For example, during chlorine disinfection, which is an oxidation process, the wastewater

will exhibit a positive ORP. Stronger the oxidative power of chlorine, higher will be the

wastewater ORP value

•

All living matter, including microbes depend upon respiration to generate energy and

respiration involves series of chemical oxidation-reduction reactions

Bacteria grow and thrive only in speciﬁc chemical - oxidative-reductive environments

which support its inbuilt metabolic pathways

Aerobic bacteria need molecular oxygen as the terminal electron acceptor as part of its

cellular respiration proces. Bacteria adapted to exist in an environment where molecular

oxygen is not present (anoxic and anaerobic), rely on electron acceptors such as NO⁻₃(den-

itriﬁcation), SO⁻₄(sulﬁde formation) and carbon (methane formers in anaerobic digestion)

Aerobic bacteria responsible for cBOD removal in the secondary treatment process would

be inhibited or wiped-out if the wastewater oxidation potential dropped and become re-

ductive. Likewise, if the wastewater in the sewer pipes which is normally of reductive

(negative ORP) was to become oxidative because of aeration (dissolving oxygen) it would

cease the hydrogen sluﬁde activity of the anaerobic bacteria

•

Typical Wastewater Process ORPs

Clorine disinfection

+650 to +700 mV

+100 to +350 mV

+20 to +250mV

+50 to +250 mV

+50 to -50 mV

- 200 mV

Nitriﬁcation

Biological phosphorous removal

Activated sludge cBOD degradation with free molecular oxygen

Denitriﬁcation

Inﬂuent wastewater

Sulﬁde formation

-50 to -250 mV

-100 to -225 mV

-100 to -250 mV

-75 to -400 mV

Anaerobic Digestion: Acid formation (Acidogenesis)

Biological phosphorous removal

Anaerobic Digestion: Methane production (Methanogenesis)

4.2 Wastewater Properties and Parameters

4.2.3 Chlorine Residual

• Chlorine in one form or another is the principal disinfecting agent employed.

•

An important additional advantage over some other disinfectants is that chlorine leaves a

disinfectant residual that assists in preventing recontamination during distribution, trans-

port, and household storage of water.

•

The absence of a chlorine residual in the distribution system may, in certain circumstances,

indicate the possibility of post-treatment contamination.

• Three types of chlorine residual may be measured:

1. Free chlorine - the most reactive form, which is the chlorine present as hypochlorite

(OCl⁻), hypochlorous (HOCl) or a combination of the two.

2. Combined chlorine - the less reactive but more persistent form, consisting of chlo-

rine that is combined with ammonia, nitrogen, or nitrogenous compounds (chlo-

ramines). This is the amount of chlorine that has reacted with nitrates and is unavail-

able for disinfection.

3. Total chlorine - which is the sum of the free and combined chlorine residuals.

•

Free chlorine is unstable in aqueous solution, and the chlorine content of water samples

may decrease rapidly, particularly at warm temperatures. Also, exposure to strong light or

agitation will accelerate the rate of loss of free chlorine. Water samples should therefore be

analyzed for free chlorine immediately on sampling and not stored for later testing.

• DPD - N,N-diethyl-p-phenylenediamine, is a commonly used method of measuring the

chlorine residual in water. DPD reacts directly with disinfectants (e.g. chlorine, chlo-

ramines etc.) to produce a pink colored solution. The intensity of this colored solution is

proportional to the concentration of disinfectant in the sample.

• Amperometric sensors are also used for chlorine measurements.

• Advantages of amperometric method include:

– Chemical reagents are not required

–

Sensors are relatively free from interference from color, turbidity, and interference

from iron, manganese, nitrate and chromates present in the sample.

4.2.4 Alkalinity

• Alkalinity is the ability of a water to neutralize acids.

•

During certain wastewater treatment processes including anaerobic digestion, acids are

generated as a result of microbiological activity. The bacteria and other biological entities

which play an active role in wastewater treatment are most effective at a neutral to slightly

alkaline pH of 7 to 8. In order to maintain these optimal pH conditions for biological

activity there must be sufﬁcient alkalinity present in the wastewater to neutralize acids

generated by the active biomass.

•

This ability to maintain the proper pH in the wastewater as it undergoes treatment is the

reason why alkalinity is so important to the wastewater industry.

Chapter 4. Constituents Properties and Analysis

•

The alkalinity is due to the presence of acid neutralizing bases in the water including the

hydroxyl (OH⁻), carbonate (CO₃⁻) and bicarbonate (HCO₃⁻) ions. These ions are of

mineral origin and are also formed from carbon dioxide which comes from the atmosphere

and from the microbial decomposition of organic material. The resistance to pH change of

the water will continue until all the alkalinity contributing ions are neutralized.

The pH of a water serves as a guide to the types of alkalinity present in the water but is

unrelated to the alkalinity content of a water. Important Note: Alkalinity is a measure of

the ability to neutralize acids whereas a solution is termed alkaline (or basic) if its pH

greater than 7.

• Alkalinity is expressed as milligrams per liter of CaCO₃

4.2.5 Dissolved Oxygen

•

Dissolved oxygen (DO) is the concentration of oxygen dissolved in the wastewater sample

and is typically measured in the ﬁeld using an amperometric DO probe which utilizes an

oxygen permeable membrane that enables a chemical reduction reaction, which produces

an electrical signal to capture the DO concentration value.

The presence of oxygen indicates an aerobic environment where dissolved, free oxygen is

available for aerobic microorganisms to live, BOD removal in the activated sludge process

occurs as a result of the activity of aerobic bacteria. The absence of DO indicates that the

environment or condition is either anoxic or anaerobic.

In an anoxic environment, free oxygen is not present, but oxygen is available from

its combined forms - nitrate (NO₃⁻) and sulfate (SO₄⁻) for the the consumption of

microorganisms. Example of an anoxic process is denitriﬁcation. In denitriﬁcation, the

anoxic bacteria in the presence of food (cBOD) consume the combined oxygen in nitrates

(NO₃⁻) and convert it to nitrogen gas.

•

The complete absence of oxygen including free and combined oxygen is an anaerobic

environment.

Microorganisms are termed as obligate aerobes if they cannot survive without free oxygen.

Facultative aerobes are microorganisms which can survive in both aerobic and anaerobic

environments.

4.2.6 Microbiological testing and monitoring

Microbes play a critical role in wastewater treatment.

•

Heterotrohic (organisms that consume organic material) microbes are responsible for the

biological wastewater treatment processes - secondary treatment process, digestion and

nutrient removal; and

• Pathogens - agents that cause disease are present in wastewater efﬂuent.

Microbiological testing and monitoring is conducted as part of the wastewater treatment typically

for the following:

1. Microbiological testing related to monitoring and troubleshooting biological wastewater

4.2 Wastewater Properties and Parameters

treatment

Microbes involved in biological wastewater treatment processes include:

•

Fungi - Filamentous fungi occasionally bloom in activated sludge processes due to

low pH or nutrient deﬁciency and cause problems with the settleability.

•

Protozoa - Protozoas play a important role in the secondary treatment process. Com-

mon protozoas in the activated sludge process include:

– Amoeba

– Flagellate

– Cilliate

• Rotifers

• Nematodes

•

Bacteria - Bacteria is the predominant microorganism responsible for the biological

wastewater water treatment.

The effectiveness of the biological wastewater treatment processes is primarily due to

the presence of a microbial ecosystem with a right balance of populations of different

microbial species.

•

Methods used for monitoring the microbial composition include direct monitoring us-

ing a light microscope to see which and how many of the different microbial species

are present - typically used for activated sludge process.

•

Indirect method includes monitoring other parameters such as pH and alkalinity

which are inﬂuenced by microbiological activity.

The microbial monitoring ensures process stability and helps identify potential pro-

cess upset conditions caused by changes to the microbial population due to other

external factors - toxicty, organic loading, temperature etc.

2. Microbiological testing related to monitoring and controlling pathogens in treated wastew-

ater efﬂuent

Pathogens in wastewater belong to the following groups:

•

Bacteria: Although, bacteria is present in large numbers in feces, pathogenic or bac-

teria are present only because of an infection and this pathogenic bacteria can poten-

tially spread the infection to other healthy individuals. Disease spread by pathogenic

bacteria include diarrhea, cholera and typhoid among many others.

Viruses: A large number of viruses may infect humans and are present in feces.

These include enteroviruses (including polioviruses), hepatitis A virus, reoviruses

and diarrhea-causing viruses (especially rotavirus).

•

Protozoa: Many species of protozoa can infect humans and cause diarrhoea and

dysentery. Girardia which casues diarrheal illness is an example of a protozoan

pathogen

•

Helminths: These are parasitic worms that can infect humans and are transmitted to

others through its eggs or larval forms

As one of the main reasons for treating wastewater is to protect public health, micro-

Chapter 4. Constituents Properties and Analysis

biological/pathogen testing of the wastewater efﬂuent and the surface water impacted

by the wastewater discharge is conducted to meet the requirements of a wastewater

discharge permit, to monitor the pathogen impact of treated wastewater discharge and

assess the level of contamination of a public body of water.

•

The bacteriological tests involves detection and quantiﬁcation of one or more of the

following bacteria: total coliforms, fecal coliforms, E. Coli, and Enterococcus.

–

The main reason why these bacteria such as coliforms and enterococcus are used

as it is not practical to detect and quantify all pathogens associated with wastewater.

These selected bacteria originate from feces and indicate fecal contamination

and thus serve as an indicator organisms for pathogens of wastewater origin.

Also, they are abundant, potentially less harmful, and easy to detect. E. coli has

been shown to be a better predictor of the potential for impacts to human health

and therefore many newer wastewater discharge permits require E. Coli testing

in lieu of fecal coliform testing requirements.

•

The microbiological test sample is always collected as a grab in a clean, sterile

borosilicate glass or plastic bottle containing sodium thiosulfate.

–

Sodium thiosulfate is added to remove residual chlorine which will kill col-

iforms during transit.

If the sample is not preserved or maintained under proper conditions until the

test is conducted in the laboratory, the test would provide erroneous results.

Samples must be refrigerated if they cannot be analyzed within 1 hour of col-

lection and must be handled with care to prevent contamination and adverse

conditions such as prolonged exposure to direct sunlight.

The maximum holding time for state or federal permit reporting purposes is 6

hours.

–

•

As it is not possible to exactly quantify the number of bacteria present, a statistical

based - Most Probable Number (MPN) approach is utilized. The methods for wastew-

ater bacteriological tests include: multiple-tube fermentation technique, membrane

ﬁltration and quanti-tray testing.

4.2.7 Speciﬁc Gravity

• Speciﬁc gravity is a term to express the weight of a solution with respect to that of water

• Water weighs 1 kg/L or 8.34 lbs/gallon or 62.4 lbs/ft³

•

A solution with a speciﬁc gravity of 1.2 will weigh 1.2 times the same volume of water. 1

L of that solution will weigh ( 1.2 kg )/L or ( 1.2*8.34=10lbs )/gallon.

Typically wastewater and the associated unthickened sludge, for all practical purposes is

assumed to have a speciﬁc gravity of 1 - implying 8.34 lbs/gallon.

Speciﬁc gravity is typically used for calculations related to chemicals used in wastewater

treatment.

4.3 Wastewater Sampling

•

Field or laboratory measurement of a certain parameter is critical in wastewater treatment

operations to obtain information about wastewater characteristics in order to either charac-

terize a wastewater stream, or to monitor a treatment process or for permit compliance.

A sample is a small part of the whole representing the whole. Thus, a sample needs to be

such that it truly represents the entire population – which in a wastewater operations could

be either a wastewater stream, wastewater solids or a chemical used.

4.3.1 Sampling Methods

Grab Samples

• A grab sample is a sample collected at a speciﬁc spot at a site over a short period of time.

•

Grab sampling allows for instantaneous analysis of parameters such as pH, dissolved

oxygen, chlorine residual, temperature and other parameters which change rapidly with

time.

• A grab sample represents a snapshot of space and time of a process stream.

Composite Samples

•

A composite sample is a collection of discrete samples are combined over a certain period

or space and therefore represent the average performance of a wastewater treatment plant

or a process during the collection period.

• Composite sampling can be either based on:

1. constant time interval (time proportioned sampling)

2. constant wastewater volume interval (ﬂow-proportioned sampling), and

3. treatment process space - includes samples taken at different depths

•

Composite samples are typically collected using automated samplers which can be pro-

grammed to collect samples at pre-established time intervals – for time proportional sam-

pling.

•

Time and space composite samples are collected by adding equal volumes of samples

collected from different times or locations.

Flow proportional composite samples comprise of volume of each subsample based on

ﬂow.

Grab Sampling Using a Long Handle Dipper

Automated Sampler

Chapter 4. Constituents Properties and Analysis

Sampling Precautions and Protocols

•

Samples should represent the major portion of the process or the process stream and

should be taken from places where the mixing is thorough, avoiding dead spots and ar-

eas of heavier or lighter loadings.

•

The collected sample is invariably exposed to conditions very different from the original

source and is subject to change due to chemical and microbiological activity.

Thus, in order to ensure integrity of sample, sample preservation techniques speciﬁc to the

analysis to be performed is needed.

–

The preservation technique should not only allow for stabilizing the parameter to be

analyzed, it should also not interfere with the analyses.

–

The common preservation techniques involve use of proper containers, temperature

control, addition of chemical preservatives, and observance of the recommended

maximum sample holding time.

Bacteriological Sampling

• Always collected as a grab

•

A clean, sterile borosilicate glass or plastic bottle containing sodium thiosulfate is used.

Sodium thiosulfate is added to remove residual chlorine which will kill coliforms during

transit. If the sample is not preserved or maintained under proper conditions until the test

is conducted in the laboratory, the test would provide erroneous results

• Samples must be refrigerated if they cannot be analyzed within 1 hour of collection

•

Samples must be handled with care to prevent contamination and adverse conditions such

as prolonged exposure to direct sunlight

• Maximum holding time for state or federal permit reporting purposes is 6 hours

4.3.2 Data Reporting

•

Arithmetic mean is typically calculated for reporting data where multiple samples have

been collected and analyzed for the same process stream at different times and for report-

ing average value over a certain time period – daily, monthly etc.

Arithmetic mean mathematically is calculated by adding all the result values and dividing

by the total number of data points.

Mathematically the arithmetic mean is represented as:

_i=1xⁱx₁+x₂+x₃...x_n

∑

x¯ =

For example:

Arithmetic mean of the following set of data points: 200, 304, 250, 400 is calculated as:

200+302+250+400

Arithmetic Mean =

= 288

For data sets for analysis such as fecal coliform could include values which vary by several

4.4 Laboratory Analysis

orders of magnitudes, using the arithmetic mean to report the average value is not appropriate as

the lower or higher values would bias the calculated mean.

For example, consider a data set with values: 260, 300, 500, 5,000, 320 and 200.

260+300+500+5,000+320+200

The arithmetic mean =

= 3,444

Here the 5000 value completely skews the arithmetic mean.

Therefore, for such tests, the geometric mean calculation is used for reporting the average value.

Mathematically a geometric mean is represented as:

√

ⁿa₁∗a₂∗a₃...a_n

∏

i=a

Calculation method:

1. Find the product of all the data points (analogous to ﬁrst calculating the sum of all the data

points when calculating the arithmetic mean)

260*300*500*5,000*320*200 = 12,480,000,000,000,000

2. Raise the product to the inverse of the number of data points

(*Using the power function of a scientiﬁc calculator)

Here n (# data points) = 6 =⇒ geometric mean = (12,480,000,000,000,000) = 482

4.4 Laboratory Analysis

4.4.1 BOD Analysis

•

The Biochemical Oxygen Demand (BOD) test estimates the amount of biodegradable ma-

terial present by measuring the amount of oxygen used by the bacteria to break down the

organic waste in the sample incubated at 20 deg. C over a ﬁve-day period . The BOD test

provides an indication on the strength of wastewater in terms of how much oxygen could

be depleted if that wastewater was introduced into another receiving water. Complete sta-

bilization of a sample may require a period of incubation too long for practical purposes;

therefore, 5 days has been accepted as the standard incubation period.

•

As the regular BOD test includes estimation of oxygen nitrifying bacteria consumes in the

process of converting inorganic forms of ammonia and nitrogen to nitrite and nitrate, its

value represents oxygen used for removing both, organic material and nitrogenous mat-

ter. As this BOD value does not quite represent the organic strength of the wastewater,

the normal BOD test is modiﬁed by introducing a chemical inhibitor - 3 mg of 2-chloro-

6-(trichloro methyl) pyridine (TCMP), which suppresses the growth of the nitrogenous

bacteria so that the resultant BOD measured represents the oxygen depletion associated

with the depletion of the organic matter only. This is the Carbonaceous biochemical oxy-

gen demand or cBOD.

Chapter 4. Constituents Properties and Analysis

Thus tBOD = nBOD + cBOD

•

Wastewater BOD measurement involves testing a sample set consisting of several sample

dilutions along with a "Blank". "Blank" is a sample with only the dilution water with no

wastewater added.

The dilutions are made based upon the expected BOD concentration of the sample. Using

the ﬁnal dilution volume of 300 ml, the initial sample volume can be estimated using the

formula:

ꢀ

ꢁi

Oxygen Depletion

ꢀ

ꢁ

Sample Volume(ml) =

∗300 ml

Anticipated BOD

•

For example, if testing an inﬂuent wastewater BOD with an expected BOD value of

250 mg/l, a range of sample volumes for dilution around sample volume of

∗

250

300 ml=5ml.

The data obtained for each of the dilutions after the 5-day incubation period must meet the

following criteria for the sample value to be acceptable for calculating the BOD.

1. A residual DO of at least 1 mg/L,

2. A DO depletion of at least 2 mg/L

•

Additionally, the whole sample set is rejected if the Blank shows an oxygen depletion of

>0.2mg/l.

• BOD is calculated for each sample dilution value using the following formula:

ꢀ

ꢁ

Initial DO−DO Day 5

∗300 ml

Sample Volume (ml)

BOD

4.4.2 Wastewater solids

Total (TSS) and Volatile (VSS)

• A known volume of wastewater sample is ﬁltered through a pre-weighed ﬁlter paper

• The suspended solids will be retained by the ﬁlter

• The water with the dissolved solids will pass through the ﬁlter

• The ﬁlter paper with the ﬁlter solids is rinsed with distilled water to remove

4.4 Laboratory Analysis

• The ﬁlter paper with the solids is dried in the oven and then weighed

•

The difference between the weight of the dried ﬁlter paper with the solids and the pre-

weighed ﬁlter paper, measured in mg, will be the suspended solids in: mg per the original

quantity of wastewater sample taken. This value can be converted to give the suspended

solids content in mg/l

• A ﬁlter paper with the dried solids is incinerated in a mufﬂer furnace

• The difference in the weight of the solids, before and after incineration is the ﬁxed solids

•

The difference between the weight of the solids before incineration and the ﬁxed solids is

the volatile solids

ꢀ

weight of solids gms 1000 ml 1000 mg

ꢀ

Total Suspended Solids - TSS TSS

∗

ꢀ

gms

volume of sample ml

ꢀ

weight of filter paper with dried solids−weight of filter paper

volume of sample (ml)

∗1,000,000

Volatile Suspended Solids - VSS

mg weight of volatile solids gms 1000 ml 1000 mg

ꢀ

VSS

∗

ꢀ

gms

volume of sample ml

ꢀ

wt. of filter paper with dried solids−wt. of filter paper incinerated residue

volume of sample (ml)

∗

1,000,000

(

weight (gms) of volatile solids gms volatile solids 100 gms total solids

(

VSS(%) =

∗

(

100 gms total solids

gms total solids

(

wt. o f filter paper with dried solids−wt. of filter paper incinerated residue

∗100

wt. o f filter paper with dried solids−wt. of filter paper

Wastewater and Sludge Total & Volatile Solids

•

A certain quantity of wastewater (by volume) or sludge (by weight) is taken in a pre-weighed dish

and weighed. Note: the sample is not ﬁltered.

• The dish with the sample is dried in an oven

•

The difference in the weight of the pre-weighed dish from that of the dish with the dried sample is

the total solids

• The dried solids are incinerated in a mufﬂer furnace

• The difference in the weight of the solids, before and after incineration is the ﬁxed solids

• The difference between the ﬁxed solids and the total solids is the volatile solids

•

Total solids of a sludge sample is reported as a % of the sludge weight. A 7% sludge has 7 lbs of

Chapter 4. Constituents Properties and Analysis

solids for every 100 lbs of sludge.

For sludge samples, volatile solids is typically reported as the volatile solids fraction in % of the total solids

content of the sludge. For example, if a 8% sludge (i.e sludge which has 8% TS or 80,000mg/l solids), is

reported to have 70% volatile, it means that 70% of the total solids - 0.7*8%=5.6% or 56,000mg/l is the

sludge volatile solids content. 70% volatile does not meet the sludge has 700,000mg/l volatile solids

weight o f solids (gms)

100 gms o f sample

gms solids

Total Solids - TS TS(%) =

∗100

gms sample

weight o f cruicible with dried solids−weight ofcruicible

weight o f cruicible with sample−weight ofcruicible

∗100

weight of volatile solids (gms) gms volatile solids

Total Volatile Solids - VS VS(%) =

∗100

gms total solids

100 gms of total solids

wt. o f cruicible with dried solids−wt. of cruicible incinerated residue

wt. o f cruicible with dried solids−wt. of cruicible

∗100

4.4 Laboratory Analysis

Sample BOD and solids analysis math problems

1. BOD tests are run on the ﬁnal efﬂuent from an activated sludge plant with and without the use of

a "nitriﬁcation inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The

raw data for these tests are presented below. What percentage of the average total BOD is the

average nBOD?

Sample Volume, ml

Blank

Initial DO, mg/l

Final DO, mg/l

9.0

6.9

8.9

4.8

8.8

2.5

9.1

1.1

9.1

9.0

BOD Test with "inhibitor" added (cBOD)

Sample Volume, ml

Blank

Initial DO, mg/l

Final DO, mg/l

8.9

7.5

8.9

6.2

9.0

5.0

9.0

3.3

9.1

9.0

Solution:

Blanks for both tBOD and cBOD are both <=0.2mg/l - thus sample sets are acceptable

Sample Volume, ml

tBOD Diff., mg/l

tBOD, mg/l

2.1

4.1

6.3

2.1*300/10

=63.0

4.1*300/20

= 61.5

6.3*300/30

= 63.0

8.0*300/40

= 60.0

cBOD Diff., mg/l

cBOD, mg/l

1.4

2.7

4.0

5.7

Reject

2.7*300/20

= 40.5

4.0*300/30

= 40

5.7*300/40

= 42.75

Depletion < 2

tBOD(avg) = (63+61.5+63+60)/4 = 61.9

cBOD(avg) = (40.5+40+42.75)/3 = 41.1

nBOD = tBOD - cBOD =⇒ nBOD = 61.9-41.1=20.8 =⇒ nBOD(%)=20.8/61.9*100= 33.6%

Chapter 4. Constituents Properties and Analysis

2. Calculate percent total solids and percent volatile solids of a sludge sample given the following

data:

Weight of dish

104.55 gms

199.95 gms

108.34 gms

106.37 gms

Weight of dish and wet sludge

Weight of dish and dry sludge

Weight of dish and ash

Solution:

Weight of dish=104.55 gms

Weight of dish and wet sludge=199.95 gms

Weight of dish and ash = 106.37 gms

=⇒ Weight o f sludge = 199.95−104.55 = 95.40 gms

=⇒ Weight o f dry sludge (solids) = 108.34−104.55 = 3.79 gms

=⇒ Weight o f volatile solids = 108.34−106.37 = 1.97 gms

(

3.79 gms solids 100 gms sludge

(

gms solids

100 gms sludge 95.40 gm(s sludge 100 gms sludge

Total solids(TS%) =

∗

= 3.97%

= 52.0%

(

1.97 gms volatile solids 100 gms total solids

(

Total volatile solids(VS%) =

∗

(

3.79

100 gms total solids

gms total solids

(

4.4.3 Bacteriological Enumeration

•

Involves bacteriological testing of the wastewater efﬂuent and the surface water impacted by the

wastewater discharge

• Conducted in-order to:

1. Meet the requirements of a wastewater discharge permit

2. Monitor the pathogen impact of treated wastewater discharge

3. Assess the level of contamination of a public body of water

4. Bacteriological tests involves detection and quantiﬁcation of one or more of the following

bacteria: total coliforms, fecal coliforms, E. Coli, and Enterococci.

Wastewater Bacteria

5. In wastewater, fecal coliforms originate in the intestines of warm-blooded animals. Aerobic

bacteria including coliforms partake in the metabolization of the organic matter as part of the

secondary treatment process

6. Fecal coliforms are seldom pathogenic under normal circumstances and are easily cultured,

their presence indicates the potential presence of pathogens

4.4 Laboratory Analysis

The reason why these bacteria such as coliforms and enterococcus are used:

(a) It is not practical to detect and quantify all pathogens associated with wastewater

(b) These bacteria originate from feces and indicate fecal contamination and thus serve as

an indicator organisms for pathogens of wastewater origin

– abundant

– potentially less harmful, and

– easy to detect

(d) E. coli has been shown to be a better predictor of the potential for impacts to human

health and therefore many newer wastewater discharge permits require E. Coli testing in

lieu of fecal coliform testing requirements.

Bacteriological Testing Methods

The methods for wastewater bacteriological tests include: multiple-tube fermentation (MTF) technique,

membrane ﬁltration (MF) and quanti-tray testing. When using the MTF and MF methods, it is not possi-

ble to exactly quantify the number of bacteria present, a statistical based - Most Probable Number (MPN)

approach is utilized

The Multiple-Tube Fermentation (MTF) technique

This involves adding three volumes – 10 ml, 1 ml and 0.1 ml of the sample, each to a set of ﬁve tubes

containing Lauryl Tryptose broth and an inverted tube (Durham tube), followed by incubating the tubes at

for a speciﬁed time. The Lauryl Tryptose broth produces color and/or turbidity change due to the growth

of the target bacteria and the inverted tube collects the gas produced by the bacterial respiration. At the

end of the process, the number of tubes showing bacterial growth are counted for each volume of sample

and using this information the concentrations of organisms in the original sample are established using

Statistical Tables. The test is conducted in three parts – presumptive, conﬁrmative and completed. A

schematic of the MTF used for quantifying total coliforms and fecal coliforms is provided below.

Wastewater Sample

Make required serial dilutions (See Note 1)

Example: Expected coliform value range 5000 – 50000.

First dilution (1:10): 11 ml sample + 99 ml dilution water

Second dilution (1:100): 11 ml of ﬁrst dilution + 99 ml dilution water

Third dilution (1:1000): 11 ml of second dilution water + 99 ml dilution water

Setup 15 tubes with Lauryl Tryptose Broth and an inverted vial (Durham tube)

5 tubes each for 10 ml, 1 ml and 0.1ml sample

10 ml

1 ml

0.1 ml

Incubate 24 hrs – check tubes for positive results marked by : 1) broth turning cloudy, and 2) gas collection in

the inverted tube. If no gas collection or media clouding is observed, incubate for another 24 hrs and then

check again to see if the tubes are positive

- + - - -

+ - - + +

10 ml

1 ml

0.1 ml

Inoculate each positive into bacteria speciﬁc broths – EC broth for fecal coliforms and Brilliant Green Bile (BGB)

broth for total coliforms with inverted tubes. Incubate for upto 24 hrs for fecal coliforms and 24 hours for total

coliforms and observe for positive results

EC Broth (for fecal coliform}

Note: EC Broth with MUG is used for E. Coli

BGB Broth (for total coliform)

Count number of conﬁrmed positives for each set of the three sample volumes

Example above: 10 ml – 3, 1 ml – 1, 0.1 ml - 1

Establish the MPN using the statistical MPN Table

Example 3-1-1 From Table - MPN = 14 MPN/100 ml. Given 1:1000 dilution, MPN = 14 * 1000 = 14,000 MPN/100 ml

If after 24 hours, from the gram staining of the colonies

Use colonies to innoculate an

agar slant and nutrient broth

from the agar slant, nonsporing gram negative bacteria

can be identiﬁed and the lactose broth shows gas for-

mation, the completed test is deemed positive

Streak agar plates & incubate

Chapter 4. Constituents Properties and Analysis

The Membrane Filtration (MF) method

This is a faster way to estimate bacterial populations in water. In this method, an appropriate sample

volume is passed through a membrane ﬁlter with a pore size small enough (0.45 micron) to retain the

bacteria present. The ﬁlter is placed on an absorbent pad (in a petri dish) saturated with a culture medium

that is selective for coliform growth. The petri dish containing the ﬁlter and pad is incubated, upside down,

for 24 hours at the appropriate temperature. After incubation, the colonies that have grown are identiﬁed

and counted using a low power microscope. A MUG medium is used for E- Coli. If E. Coli is present, it

will make the MUG ﬂuorescent when viewed in UV light.

4.4 Laboratory Analysis

Quanti-trays tests

This test used for the detection and quantiﬁcation of speciﬁc microorganisms is being used increasingly

mainly because it is a quicker test than the MTF. Colilert and Enterolert are the quanti tray based tests for

E. Coli and Enterococcus. This method involve the use of speciﬁc enzymes and overcomes the drawbacks

of the MTF which include false positives and negatives due to the more generic nature of the media used

Chapter Assessment

1. Positive ORP indicates the presence of an

environment

is the predominant microorganism responsible for biological wastewater treatment pro-

cesses

3. MPN is a statistics based method to estimate the concentration of viable bacteria is wastewater and

it stands for

4. Describe the three types of composite sampling and explain in your own words on how the sam-

pling would be conducted for each of those methods

5. Alkalinity helps in

activity

control when organic acids are formed as a result of microbiological

6. Nitrogen in wastewater is typically present as

and

7. Volatile solids is determined by incinerating the solids at

deg. C in a

8. What would be the expected BOD concentration in the wastewater if per capita (person) generation

is 0.15lb BOD per day per person and each person produces 80 gallons per day?

(Hint: The answer is going to be in mg BOD/l. So multiply lb per day per person by the inverse of

dallons per day - so that the unit is lbs per gallon and then convert that to mg/l)

9. Describe the three types of composite sampling and explain in your own words on how the sam-

pling would be conducted for each of those methods

10. tBOD =

BOD +

BOD

11. BOD stands for

12. Alkalinity helps in Nderlinehspace1cm control when organic acids are formed as a result of micro-

biological activity

13. Nitrogen in wastewater is typically present as

and

14. Volatile solids is determined by incinerating the solids at

ment]

[temp] deg. C in a

[equip-

15. How does one estimate the wastewater sample size used for BOD testing

16. Explain how is the nBOD quantiﬁed

Chapter 4. Constituents Properties and Analysis

17. Explain why coliforms and enterococcus are used for wastewater bacteriological testing

18. What is MPN and why is it used.

19. Explain the steps involved in the MTF method

20. What is the difference between volatile solids (VS) and volatile suspended solids (VSS)

21. Solids that can be used as food by microorganisms

22. Test method for determining total ammonium and organic nitrogen content:

23. Inorganic solids are also called

24. Oxygen in an anoxic environment is present as

25. Conductivity is the measure of

26. Negative ORP signiﬁes

solids

environment

27. Alkalinity helps in

control when organic acids are formed as a result of microbiological activity

28. An Imhoff cone is often used to measure the effectiveness of primary sedimentation.

a. True

*b. False

29. At a primarily domestic wastewater treatment plant, the inﬂuent wastewater BOD is always greater

than its chemical oxygen demand.

a. True

*b. False

30. Dissolved oxygen in wastewater usually is referred to as combined oxygen.

a. True

*b. False

31. Domestic wastewater generally contains only about 0.1

*a. True

b. False

32. Grab or composite samples may be used interchangeable, whichever is most convenient and safest

for all laboratory tests.

a. True

*b. False

33. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment.

*a. True

b. False

34. Pre-aeration improves settleability in primary clariﬁer

*a. True

b. False

35. Total solids are made up of dissolved and suspended solids both of which contain organic and

inorganic matter

*a. True

b. False

36. pH value less than 7 indicates an alkaline or basic condition

a. True

*b. False

37. Conductivity is a very useful test for assessing sea water intrusion into sewer lines

*a. True

b. False

38. Sample obtained for wastewater bacteriological testing is typically a composite sample

4.4 Laboratory Analysis

a. True

*b. False

39. Samples collected for the analysis of COD, BOD, and pH should be acidiﬁed for preservation.

a. True

*b. False

40. The MPN test is used to measure pathogen concentrations in wastewater.

a. True

*b. False

41. The total solids in wastewater would be the combination of the ﬁxed solids and the settleable solids.

a. True

*b. False

42. Wastewater with a pH of 2.0 to 4.0 would be too "basic" for biological treatment.

a. True

*b. False

43. Match the measurement units for each of the following:

44. The MPN test measures the number of pathogens in a wastewater sample

a. True

*b. False

45. The values of organic matter present in wastewater as measured by BOD and COD tests are typi-

cally almost identical

a. True

*b. False

46. In order to obtain valid results in coliform testing, the sample must be dechlorinated at the time of

its collection.

*a. True

b. False

47. A ﬂow of 100 gallons per capita per day is often used for estimating ﬂow into a wastewater treat-

ment plant.

*a. True

b. False

48. An Imhoff cone is used to measure settleable solids in units of mg/l.

a. True

*b. False

49. A standardized method exists for the measurement of ﬂoatable solids.

a. True

*b. False

50. In the test for coliform bacteria, samples are incubated for 5 days at 20 deg. C.

a. True

*b. False

51. Coliform group organisms are found only in wastewater.

a. True

*b. False

52. Where highly colored samples are involved, the determination of pH by use of a pH meter rather

than by color-comparison is

preferred.

Chapter 4. Constituents Properties and Analysis

*a. True

b. False

53. What comes into a treatment plant must go out. This is the basis of the solids balance concept.

*a. True

b. False

54. Receiving water measurements are used to determine the effect of the plant’s waste discharge on

the receiving waters; therefore, it is necessary to measure both stream and plant

efﬂuent characteristics.

*a. True

b. False

55. There are two types of samples that may be collected. One is called an integrated sample, and the

other is referred to as a composite.

a. True

*b. False

56. Coliform testing is typically conducted on a grab sample

*a. True

b. False

57. Typical domestic wastewater BOD content is about 2000mg/l

a. True

*b. False

58. BOD is a measure of the organic content of wastewater

*a. True

b. False

59. ORP measurements can be used for controlling the disinfection process

*a. True

b. False

60. MPN is used for enumerating wastewater bacteria and it stands for Measured Pathogen Number

a. True

*b. False

61. For measuring dissolved oxygen in wastewater, it is advisable to use a composite sample

*a. True

b. False

62. A sample to be used for pH measurements should be preserved by the addition of an acid prior to

its analysis in the laboratory

a. True

*b. False

63. BOD is a measure of the organic content of wastewater and it stands for Biological Oxidation

Demand

a. True

*b. False

64. Conductivity is measured in the units of millivolts using an electrochemical probe

a. True

*b. False

65. Conductivity measurements provide an indirect way to measure total solids present in wastewater

a. True

4.4 Laboratory Analysis

*b. False

66. Typical wastewater TSS concentrations are in the 2,000-2,500 mg/l range

a. True

*b. False

67. A 24-hr ﬂow proportioned sample may be collected for a fecal coliform test

a. True

*b. False

68. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment.

*a. True

b. False

69. The laboratory measurement of volatile solids is a fair approximation of the organic content of the

wastewater.

*a. True

b. False

70. BOD5 and SS are both used to measure the strength of wastewater

*a. True

b. False

71. Pre-aeration improves settleability in primary clariﬁer

*a. True

b. False

72. pH value less than 7 indicates an alkaline or basic condition

a. True

*b. False

Chapter 4. Constituents Properties and Analysis

73. A BOD test is run on a secondary efﬂuent. The data for this test are given below. Assuming that the

average BOD test result for an “inhibitor” on this same secondary efﬂuent was 22 mg/L (cBOD),

calculate the oxygen demand caused by nitriﬁcation. What percentage of the total BOD is the

nitrogenous oxygen demand? On April 04, 2009 Exam

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

3.5

8.9

1.2

Blank

9.2

9.0

• Solution:

Dif ference,mg/l

5.5

5.5∗300/50 = 33

7.7

7.7∗300/75 = 30.8

9.0

tBOD,mg/l

(33+30.8)

Average:

= 31.9^mg

tBOD = cBOD+nBOD =⇒ 31.9 = 22 + nBOD =⇒ nBOD = 9.9

%nBOD = 9.9/31.9∗100 = 31%

4.4 Laboratory Analysis

74. BOD tests are run on the ﬁnal efﬂuent from an activated sludge plant with and without the use of a

"nitriﬁcation inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The raw

data for these tests are presented below. What is the average NITROGENOUS BOD (NBOD)?

Exam on April 04, 2009

BOD Test without "inhibitor" (tBOD)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

5.1

8.7

1.2

Blank

9.1

9.0

BOD Test with "inhibitor" added (cBOD)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

6.5

8.7

3.5

Blank

9.1

9.0

• Solution:

tBOD

Dif f.,mg/l

3.9

7.5

0.1

tBOD,mg/l

cBOD

3.9∗300/30 = 39

7.5∗300/60 = 37.5

Dif f.,mg/l

tBOD,mg/l

2.5

5.2

2.5∗300/30 = 25

5.2∗300/60 = 26

Average:

(39+37.5)

tBOD:

cBOD:

= 38.3^mg

(25+26)

= 25.5^mg

tBOD = cBOD+nBOD =⇒ 38.3 = 25.5 + nBOD =⇒ nBOD = 12.8

Chapter 4. Constituents Properties and Analysis

75. Calculate the TSS of the secondary efﬂuent given the following:

Sample volume

41 ml

Tare weight of ﬁlter sample

Filter + dried residue

1.4604 gm

1.4722 gm

76. The technician quickly pours 41 ml of well mixed inﬂuent into the ﬁlter funnel. The tare weight of

the ﬁlter is 1.4604 gm. After rinsing, drying, cooling, and weighing, the ﬁrst dry weight is 1.4722

gm. The ﬁlter is returned to the oven and dried, cooled, and weighed. The second dry weight is

1.4700 gm. Calculate the TSS.

• Solution

(1.4722−1.4604)gm TSS

1000ml

∗1000

∗

= 288

41ml

4.4 Laboratory Analysis

77. BOD tests are run on the ﬁnal efﬂuent from an activated sludge plant with and without the use of

a "nitriﬁcation inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The

raw data for these tests are presented below. What percentage of the average total BOD is the

average nBOD? (10 points)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

6.9

8.9

4.8

8.8

2.5

9.1

1.1

Blank

9.1

9.0

BOD Test with "inhibitor" added (cBOD)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

8.9

7.5

8.9

6.2

9.0

5.0

9.0

3.3

Blank

9.1

9.0

Solution:

tBODDif f.,mg/l

tBOD,mg/l

2.1

4.1

6.3

2.1∗300/10 = 63.0 4.1∗300/20 = 61.5 6.3∗300/30 = 63.0 8.0∗300/40 = 60.0

1.4 2.7 4.0 5.7

4.0∗300/30 = 40 5.7∗300/40 = 42.8

cBODDif f.,mg/l

cBOD,mg/l [Discard −dif f. < 2] 2.7∗300/20 = 40.5

tBOD(avg) = 63+61.5+63+60 = 61.9

cBOD(avg) = 40.5+40+42.75 = 41.1

nBOD = tBOD - cBOD =⇒ nBOD = 61.9-41.1=20.8 =⇒ nBOD(%)=20.8/61.9*100= 33.6%

Chapter 4. Constituents Properties and Analysis

78. BOD tests are run on the ﬁnal efﬂuent from an activated sludge plant with and without the use of a

"nitriﬁcation inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The raw

data for these tests are presented below. What is the average NITROGENOUS BOD (NBOD)?

Exam on April 04, 2009

BOD Test without "inhibitor" (tBOD)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

5.1

8.7

1.2

Blank

9.1

9.0

BOD Test with "inhibitor" added (cBOD)

SampleVolume,ml

InitialDO,mg/l

FinalDO,mg/l

9.0

6.5

8.7

3.5

Blank

9.1

9.0

Solution:

tBOD

Dif f.,mg/l

3.9

7.5

0.1

tBOD,mg/l

cBOD

3.9∗300/30 = 39

7.5∗300/60 = 37.5

Dif f.,mg/l

tBOD,mg/l

2.5

5.2

2.5∗300/30 = 25

5.2∗300/60 = 26

Average:

(39+37.5)

tBOD:

cBOD:

= 38.3^mg

(25+26)

= 25.5^mg

tBOD = cBOD+nBOD =⇒ 38.3 = 25.5 + nBOD =⇒ nBOD = 12.8

4.4 Laboratory Analysis

79. Calculate percent total solids and percent volatile solids of a sludge sample given the following

data:

Weight of dish

104.55 gms

199.95 gms

108.34 gms

106.37 gms

Weight of dish and wet sludge

Weight of dish and dry sludge

Weight of dish and ash

(Answer: TS = 3.97%. VS = 52%)

Solution:

Weight of dish=104.55 gms

Weight of dish and wet sludge=199.95 gms

Weight of dish and ash = 106.37 gms

=⇒ Weight o f sludge = 199.95−104.55 = 95.40 gms

=⇒ Weight o f dry sludge (solids) = 108.34−104.55 = 3.79 gms

=⇒ Weight o f volatile solids = 108.34−106.37 = 1.97 gms

(

100 gms sludge

(

gms solids

(

3.79

(

Total solids(TS%) =

( ∗

= 3.97%

(

100 gms sludge

95.40 gms sludge 100 gms sludge

(

gms volatile solids 100 gms (total solids

1.97

(

Total volatile solids(VS%) =

∗

= 52.0%

100 gms total solids

(

3.79

(

gms total solids

(

Chapter 4. Constituents Properties and Analysis

80. What is percent volatile solids of a sludge sample given the following data:

Weight of dish

130.69 gms

249.94 gms

134.74 gms

132.05 gms

Weight of dish and wet sludge

Weight of dish and dry sludge

Weight of dish and ash

Solution:

Weight of dish=130.69 gms

Weight of dish and wet sludge=249.94 gms

Weight of dish and ash = 132.05 gms

=⇒ Weight o f sludge = 249.94−130.69 = 119.25 gms

=⇒ Weight o f dry sludge (solids) = 134.74−130.69 = 4.05 gms

=⇒ Weight o f volatile solids = 134.74−132.05 = 2.69 gms

(

100 gms sludge

(

gms solids

(

4.05

(

Total solids(TS%) =

( ∗

= 3.4%

(

100 gms sludge

119.25 gms sludge 100 gms sludge

(

gms volatile solids 100 gms (total solids

2.69

(

Total volatile solids(VS%) =

∗

= 66.4%

100 gms total solids

(

4.05

(

gms total solids

(

4.4 Laboratory Analysis

81. Products that are non-biodegradable will have

a. Same BOD

as compared with biodegradable products

*b. A lower BOD

c. A higher BOD

d. There is no relationship between BOD and biodegradability

82. Coliform bacteria are

a. Algae

b. Coagulant aids

*c. Indicators

d. Sequestering agents

83. What is percent volatile solids of a sludge sample given the following data:

Weight of dish = 130.69 gms

Weight of dish and wet sludge = 249.94 gms

Weight of dish and dry sludge = 134.74 gms

Weight of dish and ash = 132.05 gms

*a. 38.3%

b. 66.4%

c. 42.6%

d. 58.0%

84. What is percent volatile solids of a sludge sample given the following data:

Weight of dish = 130.69 gms

Weight of dish and wet sludge = 249.94 gms

Weight of dish and dry sludge = 134.74 gms

Weight of dish and ash = 132.05 gms

*a. 38.3%

b. 66.4%

c. 42.6%

d. 58.0%

85. A device called an Imhoff cone is commonly used to measure settleable solids in:

a. %

*b. mL/L

c. mg/L

d. ppm

e. SVI units

86. An aerobic treatment process is one that requires the presence of:

a. Ozone

b. organic oxygen

c. no oxygen

d. combined oxygen

*e. dissolved oxygen

87. A pH probe:

a. Can be used to measure ORP in chlorine disinfection.

b. Is often used to measure hydrogen production in wet wells.

c. Measures, in millivolts, the difference between oxidants like chlorine and reductants such as

organic matter.

Chapter 4. Constituents Properties and Analysis

*d. Measures hydrogen ion activity in wastewater.

e. Sends a 4-20 mA signal directly to a chlorine controller.

88. Sludge solids in wastewater has an average speciﬁc gravity of 1.2; this means they are

a. 12% heavier than water

*b. 20% heavier than water

c. 2% lighter than water

d. 20% lighter than water

89. How should the pH electrode be stored when not in use:

a. In a strong acid solution

b. In a strong caustic solution

c. In a safe place in a drawer

*d. In distilled water

e. In a detergent

90. In the normal Winkler test:

a. A snow white precipitate forms in direct proportion to the nitrate concentration

b. A brownish ﬂocculant precipitate is evidence that D.O. is absent

c. An endpoint is reached when a dark blue color changes to black

d. The mufﬂe furnace must be in excess of 500 deg. C before incubation

*e. A snow white precipitate forms if DO is absent

91. Organisms in wastewater that are not harmful to humans but are indicators of diseases are:

a. Pathogens

b. Viruses

*c. Coliform

d. Bacteria

92. The typical range of suspended solids in domestic inﬂuent wastewater is:

*a. 100-300 mg/L

b. 400-600 mg/L

c. 700-900 mg/L

d. 1000-12000 mg/L

93. Which of the ﬂowing statement(s) is/are true with regards to BOD

i) BOD test results are suitable for quickly establishing process efﬁciencies

ii) BOD value is always greater than the COD value of the same wastewater sample

iii) BOD is expressed in mg/ L or in ppm

iv) BOD is the measure of organic strength

v) BOD stands for biological oxygen demand

a. i) & ii)

b. i), ii) & iv)

c. i), iv) & v)

*d. iii) & iv)

e. iii), iv) & v)

94. An amperometric titrater is used to measure

a. Alkalinity

*b. Chlorine residual

c. Conductivity

d. COD.

4.4 Laboratory Analysis

95. Which of these pH readings indicates an acidic wastewater?

*a. 3

b. 7

c. 9

d. 12

96. Which of the following conditions will probably cause the greatest change in pH?

a. Buffering sample

*b. Exposing sample to atmosphere

c. Fixing sample

d. Refrigerating sample

97. Nderlinehspace1cm matter in wastewater, is normally composed of grit, sand and silt.

a. Colloidal

*b. Inorganic

c. Organic

d. Volatile

98. Which of the following characteristics would be least helpful to an operator assessing the organic

loading on his plant?

a. solids concentration

b. BOD

c. COD

*d. pH

e. nitrogen content

99. Pathogens

*a. Are bacteria or virus that cause disease.

b. Are bacteria which do not occur in water.

c. Can obtain their food supply without help.

d. Are not harmful to man.

100. Regarding the total coliform test which one of the following statements is TRUE?

a. If less than 2 coliforms per 100 mL are found in a secondary efﬂuent, we can be assured that we

have destroyed all viruses and coliforms.

b. Total coliforms are monitored in wastewater because they are generally considered to be disease

causing organisms.

c. Sodium thiosulfate should be added drop-wise after collection of a coliform sample to destroy

residual chlorine.

*d. The multiple tube fermentation method for measuring total coliforms is only a statistical esti-

mate of the coliform organism concentration.

101. Results from the Multiple-Tube Fermentation Technique for members of the Total Coliform Group

are expressed as

a. DPD.

b. MF.

c. MGD.

*d. MPN.

102. The advantage in the use of coliform organisms as an indicator lies in the following fact:

a. They are found everywhere and they grow in common bacterial media

b. They are found everywhere and are readily killed by chlorine

Chapter 4. Constituents Properties and Analysis

*c. They are predominant bacteria associated with intestinal discharges and grow on

nutrient agar forming characteristic colonies

103. The purpose of adding sodium thiosulfate to a microbiological sample bottle is to

a. Extend the allowable holding time from 6 to 30 hours.

b. React with nitrates that interfere with the MPN test.

*c. Remove any chlorine residual present.

d. To ensure sterilization of sample bottle.

104. The solids in a raw wastewater sample may be classiﬁed in several different categories. However,

one may say that all the solids in a wastewater sample may be expressed as the sum of:

a. The total of settleable solids + total dissolved solids.

*b. The total dissolved solids + total suspended solids.

c. The total of settleable solids + colloidal solids + total suspended solids

105. The term MPN is used in reference to:

a. The mass of phosphorus and nitrogen per unit of carbon

*b. The number of coliforms per unit volume of sample that is most likely to have caused the

observed results in a multiple tube test

c. The number of fecal stetococci per unit volume of sample that is most likely to have caused the

observed results in the multiple tube test

d. The result of membrane ﬁlter test

e. The standard plate count result

106. The volatile portion of suspended solids contained in normal domestic wastewater could be ex-

pected to be in the range of

a. Less than 10 b. 25-50*c. 70-80d. 90-100

107. What disease is not considered to be normally conveyed or transmitted by untreated wastewater?

a. Amoebic dysentery.

b. Hepatitis.

*c. Malaria.

d. Chlorea.

108. What test is not run on the inﬂuent?

a. BOD.

*b. Fecal coliform.

c. Suspended solids.

d. pH.

109. Which one of the following statements is TRUE regarding the standard BOD5 test?

a. Some NPDES permits specify that only CBOD is to be reported on a wastewater plant’s ﬁnal

efﬂuent. The letters CBOD refer to complete BOD (i.e the TOTAL BOD).

b. Phenylarsineoxide (PAD) is typically added to destroy nitrifying bacteria when running this test.

c. Dechlorinated secondary efﬂuents need not be seeded when set up for the BODs test.

d. Nitrate ions interfere with this test

*e. The BOD measured includes the nitrogenous BOD (nBOD).results due to the oxygen demand

exerted by certain bacteria as they oxidize ammonia to nitrate ions.

110. A composite sample will provide a(an)

a. Even color.

b. High pH.

c. Low solids sample.

4.4 Laboratory Analysis

*d. Representative sample.

111. Flow proportionate composite samples are collected because:

a. The waste characteristics are continually changing

b. The ﬂow is continually changing

*c. The ﬂow and waste characteristics are continually changing

d. This requires less time than grab samples

e. All of the above

112. Over a four-year period, the totalizing hour meter on an instrument air compressor had the follow-

ing readings at the end of each year: 1st year - 9,763; 2nd year - 13,258; 3rd year - 20,071; and 4th

year - 23,714 How many hours does the meter show the compressor ran during

the third year?

a. 349.5 hours

b. 364.3 hours

*c. 681.3 hours

d. 830.2 hours

113. Grab sample is always collected for which of the following test

a. BOD

b. TSS

*c. Coliform

d. COD

e. None of the above

114. Samples collected over several hours during the day and combined are known as:

*a. Composite samples.

b. Grab samples.

c. Deep samples.

d. Periodic samples.

115. Grab samples are considered to be representative of the

a. Average daily condition at the sample location

b. Average daily condition in the system

c. System conditions for the two hours before and after the sample was taken

*d. System condition at the time of the sample

116. Characteristics that should be measured immediately after the sample is collected are:

a. Velocity and dissolved solids

*b. Temperature, pH and DO

c. TSS and BOD

d. Hardness and alkalinity

117. A stilling well on the efﬂuent side of a wastewater facility is:

a. a chlorine injection concentrator

b. an automatic sampler for coliform counts

*c. a structure containing the ﬂoat for a ﬂow measuring device

d. dry well side of a pump station

e. a none of the above

118. The advantages of automatic sampling equipment are:

a. elimination of human error inherent in manual sampl ing

b. reduction of personnel requirements and cost

Chapter 4. Constituents Properties and Analysis

c. allows for more frequent sampling

d. collection of more representative samples

*e. all of the above

119. In collecting a sample for a chlorine residual determination of the ﬁnal efﬂuent, the most suitable

sampling point is:

a. at the site of chlorine injection

b. at the entrance to the chlorine contact chamber

c. at the midpoint of the chlorine contact chamber

d. at the exit side of the chlorinator

*e. at the point of efﬂuent discharge

120. The recommended minimum portion which should be collected for testing in sampling wastewater

is: (Assume a grab sample)

a. 10 ml

b. 50 ml

c. 10 0 ml

d. 500 ml

*e. 1,000 ml

121. A composite sample will give a(n)

a. Even color

b. High pH

c. Instantaneous sample

*d. Representative sample

122. Chlorine residual may be determined using the reagent

*a. Diethyl-p-phenylenediamine (DPD)

b. Ethylendiamine tetraacetic acid (EDTA)

c. Polychlorinated biphenyls (PCB)

d. Sodium thiosulfate (Na2S203)

123. BOD5 is the most common method to quantify the organic content in wastewater. Another method

used is:

a. Chemical oxygen demand

b. Volatile solids analysis

c. Total organic carbon

*d. Any of the above

124. The laboratory test results on domestic raw sewage were: COD = 320 mg/1 BOD = 475 mg/1. The

best interpretation would be:

a. the raw wastewater had a high grease content

b. the raw wastewater was septic

*c. the laboratory results, as reported, were in error

d. the sample was held too long before analysis

e. the glass ﬁber ﬁlters used to run the test were contaminated

125. TKN is the measure of:

a. Ammonia/Ammonium+Inorganic Nitrogen

*b. Ammonia/Ammonium+Organic Nitrogen

c. Nitrate+Nitrite+Organic Nitrogen

d. Organic Nitrogen+Inorganic Nitrogen

4.4 Laboratory Analysis

126. Volatile solids concentration of sludge with 6

*a. 42,000mg/l

b. 70,000mg/l

c. 42mg/l

d. 700mg/l

127. Wastewater solids can be categorized as:

a. Suspended+Fixed

b. Dissolved+Volatile

*c. Volatile+Fixed

d. Volatile+Settleable

128. Non-settleable solids are composed of:

a. Volatile +Dissolved

b. Dissolved+Settleable

c. Floatable + Suspended

*d. Collodial + Floatable

129. Sludge with a speciﬁc gravity of 1.1 will weigh:

a. 1.1lbs/gal

b. 8.34lbs/gal

*c. 9.17lbs/gal

d. 7.58lbs/gal

130. A composite sample will provide a(an)

a. Even color.

b. High pH.

c. Low solids sample.

*d. Representative sample.

131. Bacteria which cause diseases in man are generally called:

a. Mesophilic

b. Facultative

*c. Pathogenic

d. Coliforms

132. How should the pH electrode be stored when not in use:

a. In a strong acid solution

b. In a strong caustic solution

c. In a safe place in a drawer

*d. In distilled water

e. In a detergent

133. In the normal Winkler test:

a. A snow white precipitate forms in direct proportion to the nitrate concentration

b. A brownish ﬂocculant precipitate is evidence that D.O. is absent

c. An endpoint is reached when a dark blue color changes to black

d. The mufﬂe furnace must be in excess of 500 deg. C before incubation

*e. A snow white precipitate forms if DO is absent

134. The typical range of suspended solids in domestic inﬂuent wastewater is:

*a. 100-300 mg/L

b. 400-600 mg/L

Chapter 4. Constituents Properties and Analysis

c. 700-900 mg/L

d. 1000-12000 mg/L

135. Which of the ﬂowing statement(s) is/are true with regards to BOD

1) BOD test results are suitable for quickly establishing process efﬁciencies

2) BOD value is always greater than the COD value of the same wastewater sample

3) BOD is expressed in mg/ L or in ppm

4) BOD is the measure of organic strength

5) BOD stands for biological oxygen demand

a. 1) and 2)

b. 1), 2) & 4)

c. 1), 4) & 5)

*d. 3) & 4)

e. 3), 4) and 5)

136. Nderlinehspace1cm matter in wastewater, is normally composed of grit, sand and silt.

a. Colloidal

*b. Inorganic

c. Organic

d. Volatile

137. Pathogens

*a. Are bacteria or virus that cause disease.

b. Are bacteria which do not occur in water.

c. Can obtain their food supply without help.

d. Are not harmful to man.

138. Regarding the total coliform test which one of the following statements is TRUE?

a. If less than 2 coliforms per 100 mL are found in a secondary efﬂuent, we can be assured that we

have destroyed all viruses and coliforms.

b. Total coliforms are monitored in wastewater because they are generally considered to be disease

causing organisms.

c. Sodium thiosulfate should be added drop-wise after collection of a coliform sample to destroy

residual chlorine.

*d. The multiple tube fermentation method for measuring total coliforms is only a statistical esti-

mate of the coliform organism concentration.

139. Results from the Multiple-Tube Fermentation Technique for members of the Total Coliform Group

are expressed as

a. DPD.

b. MF.

c. MGD.

*d. MPN.

140. The advantage in the use of coliform organisms as an indicator lies in the following fact:

a. They are found everywhere and they grow in common bacterial media

b. They are found everywhere and are readily killed by chlorine

*c. They are predominant bacteria associated with intestinal discharges and grow on

nutrient agar forming characteristic colonies

141. The purpose of adding sodium thiosulfate to a microbiological sample bottle is to

a. Extend the allowable holding time from 6 to 30 hours.

4.4 Laboratory Analysis

b. React with nitrates that interfere with the MPN test.

*c. Remove any chlorine residual present.

d. To ensure sterilization of sample bottle.

142. The term MPN is used in reference to:

a. The mass of phosphorus and nitrogen per unit of carbon

*b. The number of coliforms per unit volume of sample that is most likely to have caused the

observed results in a multiple tube test

c. The number of fecal stetococci per unit volume of sample that is most likely to have caused the

observed results in the multiple tube test

d. The result of membrane ﬁlter test

e. The standard plate count result

143. Describe total coliform & fecal coliform bacteria. What is the difference between the two? De-

scribe the test procedures for both.

Response:

a Describe total coliform & fecal coliform bacteria.

•

Coliform bacteria are a broad group of bacteria found in soil, water and other environ-

ments.

• Fecal coliform are coliforms which originate in the intestines of warm-blooded animals

b What is the difference between the two?

While total coliforms are found widely in different environments, fecal coliforms are typi-

cally found in the intestines of warm blooded animals and are indicators of fecal contamina-

tion.

c Describe the test procedures for both.

The Multiple Tube Fermentation method to estimate the quantity of both these microorgan-

isms includes the following steps:

•

Conducting a presumptive test by inoculating the sample in a set of 15 tubes containing

Lauryl Tryptose Broth each containing an inverted Durham tube followed by incubation

and observation of a positive result for each tube indicated by turbidity and presence of

gas bubble.

•

Conducting a conﬁrmed test by innoculating each of the positive samples from the

presumptive test into a tube containing BGB broth - for Total Coliform and EC Broth -

for Fecal Coliform

Conducting a completed test by streaking and incubating an agar plate with positives

from the conﬁrmed test followed by innoculating and incubating an agar slant and

nutrient broth with colonies from the agar plate.

144. What is ORP Response:

Oxidation is a chemical reaction in which electrons are gained by oxidizing agent and lost by the

substance being oxidized. Electrons cause bonds to be broken in the organic matter thus destroying

it.

Oxidation potential is the direct measure in millivolts of demand for chlorine; more chlorine in-

creases oxidants potential more organic matter lowers oxidation potential.

Using ORP probe reduces amount of chlorine used for disinfection. Chlorine by-products are toxic

cause problems in efﬂuent toxicity test.

Chapter 4. Constituents Properties and Analysis

ORP (oxidation reduction potential) probes have been found to be effective in precisely controlling

chlorine disinfection as well as de-chlorination. An ORP probe measures directly the oxidation

potential and compares (his measured potential to a set point (typically 540 mV). A 4-20 milliamp

controller can be used to adjust the chlorine feed to use precisely the amount of chlorine needed for

disinfection.

145. When chlorine is added to a wastewater efﬂuent it is said to act as an oxidizing agent. Deﬁne

oxidation. Explain how oxidation might be effective in disinfection of a wastewater efﬂuent.

Response:

Oxidation is a chemical reaction in which electrons are gained by the oxidizing agent and lost by

the substance being oxidized.

These lost electrons cause bonds to be broken in the organic matter (bacteria) and thus destroy it.

146. Deﬁne oxidation potential. Explain how (his potential changes in relationship to the addition of

chlorine or the increase in organic material or bacteria in the wastewater.

Response:

Oxidation potential is a direct measure, in millivolts, of the potential for electron transfer to the

oxidizing agent (chlorine).

More chlorine increases oxidation potential.

More organic material or bacteria lowers the oxidation potential.

147. One WWTP found that it was better able to meet its efﬂuent toxicity limit when using an ORP

probe. Identify and brieﬂy explain one possible reason for this.

Response:

Using an ORP probe reduces signiﬁcantly the amount of chlorine being used for disinfection.

Chlorine by-products are often toxic and thus cause problems in the efﬂuent toxicity test (bioassay

test).

Reducing the chlorine used may make it easier to meet the toxicity limit.

148. An ORP probe must be place several minutes downstream of the point where chlorine is added in a

chlorine contact tank. Why? Identify the data and conversion factor(s) necessary to calculate where

an ORP probe is to be placed. Assume that at peak dry ﬂow (he probe is to be placed 5 minutes

downstream of the point where chlorine is added. Show the steps necessary to do this calculation.

Response:

Reactions with ammonia-nitrogen are not instantaneous. Thus, sometime between the point of

addition of chlorine and the site of the ORP probe to allow these reactions to be complete. Data

needed to calculate position of the probe: Peak Flow (MGD), the conversion factor = MGD =

1.547 cu. ft. / sec., the width of the chlorine contact chamber, the depth of the water in the chlorine

contact chamber (so that the cross-sectional area of the ﬂow can be calculated), 60 sec = 1 minute.

1. Find the velocity of ﬂow (ft/sec): Q at peak x 1.547 cu. ft./sec divided by cross-sectional area in

ft2. 2. Convert velocity (ft/sec) to ft/min = ft/sec x 60 sec/min = ft/min 3. Multiply ft/min x 5 min

to ﬁnd ft.

5. Collections

The collection system resembles a tree that branches out from the treatment plant to collect the wastewater

from individual sources.

5.1 Wastewater Collection Piping

• A lateral is the piping that connects the public sewer to the building.

• Laterals ﬂow into larger lines called mains.

• Mains carry the ﬂow into the largest lines in the system, called trunk lines.

• A trunk line is the pipe that brings water into the treatment plant.

5.2 Sanitary Sewer Systems

Sanitary sewer systems collect and convey wastewater from residential, commercial and industrial sources

to a centralized wastewater treatment facility for treatment.

5.2.1 Storm-water systems

Storm-water systems are designed solely for the conveyance of storm-waters waters directly to streams,

rivers, lakes, or the ocean.

5.2.2 Combined sewer systems

•

Combined sewer systems collect and convey sanitary sewage and urban runoff in a common piping

system.

•

Combined sewers could potentially cause serious water pollution problems during combined sewer

overﬂow (CSO) events when wet weather ﬂows exceed the sewage treatment plant capacity.

Chapter 5. Collections

Combined System

Separated System

5.3 Collections Systems Basics

•

The primary type of a collection system is a gravity system. A gravity system is so named because

the wastewater ﬂows down gradient in the sewer, driven by forces of gravity.

The collection system includes the gravity sewers, force mains, manholes, pumping equipment, and

other facilities that collect and convey the water to a wastewater treatment plant.

Sewers are generally laid at a minimum slope to ensure open channel ﬂow through the pipe at a

minimum velocity of 2.0 feet per second. The minimum velocity is required to ensure that solids do

not settle out in the sewer.

•

When the sewer lines reach a certain depth, the ﬂow must be lifted back through a lift or pump

station.

Lift stations are built whenever wastewater must be pumped to a higher altitude, whether it’s to lift

water up so that it can gravity ﬂow or to pump it over a rise or hill.

The discharge from the pump station may be to another gravity sewer at that location or through a

pressurized force main.

Key elements of lift stations include a wastewater receiving well (wet-well), pumps and piping with

associated valves.

The size of the wet well affects the operating of the station. If a wet well is too small, excessive

starting and stopping of the pump motors will occur, resulting in premature failure. If the wet well

is too large, solids will tend to settle on the bottom, blocking the pump suction line and leading to

the generation of hydrogen sulﬁde and methane.

•

The dry well is the portion of the dry well/wet well pumping station that houses the necessary

equipment required to pump the wastewater. The dry well is so named because it is isolated from

the incoming wastewater.

• Centrifugal pumps are the most common type of pump found in wastewater pumping stations.

• In the USA, wastewater generated in a typical home is about 70 gal/day/person

5.4 Collections Related Operational Issues

Inﬁltration and inﬂow (I/I) is the unwanted ﬂow into the wastewater collections systems.

5.4 Collections Related Operational Issues

5.4.1 Inﬁltration

•

Groundwater entering sanitary sewers through defective pipe joints and broken pipes is called

inﬁltration. Remember: ground ﬁlters

5.4.2 Inﬂow

•

During rainstorms or snow thaws, large volumes of water may ﬂow into the wastewater collections

systems through leaky manhole covers or combined storm-water /wastewater connections. In

addition, private residences may have roof, cellar, yard, area, or foundation drains inappropriately

connected to sanitary sewers. These ﬂows are termed as inﬂows. Remember: rain ﬂow

Implications of I/I:

• I/I decrease the efﬁciency and capacity of wastewater collection systems and treatment systems

• I/I can advance the need for capital costs to manage and treat ﬂows

•

I/I contribute to the hydraulic overloading of treatment processes, which can affect public health

and the compliance with NPDES permit requirements

•

I/I could cause Sanitary and combined sewer overﬂows(SSOs and CSOs) when wastewater ﬂow

volumes exceed the design capacity of the treatment plant

• I/I increase collection system and treatment facility operating costs

5.4.3 Odors

• Hydrogen sulﬁde (H₂S), its associated compounds and methane are generated due to microbial

activity in wastewater and the conveyance systems. The rotten egg like smell of hydrogen sulﬁde

causes public nuisance odors, poses a health hazard for collections systems workers and causes

corrosion of the system through its conversion to sulfuric acid. The hydrogen sulﬁde generation is

typically controlled by reducing the potential for septicity in wastewater through proper design of

the system - adequate velocities and adequate air space and through chemical treatment.

5.4.4 Fats, Oils and Grease (FOG)

FOG from food home, food establishments and industries affect the operation of the collection

•

system

• FOG has a tendency to accumulate in sewer pipes decreasing its wastewater carrying capacity.

• Excessive FOG accumulation may cause sewer overﬂows

Chapter Assessment

1. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

2. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to

ensure that the solids do not settle out in the sewer lines

a. True

*b. False

3. A combined sewer implies that it carries domestic and industrial wastes only

a. True

*b. False

4. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

5. Inﬂow is storm water entering into the sewer system

*a. True

b. False

6. A combined wastewater collection system handles only domestic waste and industrial waste.

a. True

*b. False

7. Fats, oils and grease accumulation in sewers can cause sewer overﬂows

*a. True

b. False

8. Storms could potentially cause sewer overﬂows in a Combined Sewer system

*a. True

b. False

9. Groundwater entering sewer collection pipes through cracks and defective pipe joints is termed as

Chapter 5. Collections

inﬂow

a. True

*b. False

10. A Combined sewer system is the one that brings the combined domestic and industrial wastewater

ﬂow to the treatment plant

a. True

*b. False

11. Inﬁltration is when groundwater enters the sewage collections systems

*a. True

b. False

12. Inﬁltration is when groundwater enters the sewage collections systems

*a. True

b. False

13. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to

ensure that the solids do not settle out in the sewer lines

a. True

*b. False

14. A combined sewer implies that it carries domestic and industrial wastes only

a. True

*b. False

15. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

16. Inﬂow is storm water entering into the sewer system

*a. True

b. False

17. A lateral is the largest sewer line which brings the wastewater to the treatment plant

a. True

*b. False

18. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

19. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to

ensure that the solids do not settle out in the sewer lines

a. True

*b. False

20. A combined sewer implies that it carries domestic and industrial wastes only

a. True

*b. False

21. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

22. Inﬂow is storm water entering into the sewer system

*a. True

b. False

5.4 Collections Related Operational Issues

23. A combined wastewater collection system handles only domestic waste and industrial waste.

a. True

*b. False

24. Fats, oils and grease accumulation in sewers can cause sewer overﬂows

*a. True

b. False

25. Storms could potentially cause sewer overﬂows in a Combined Sewer system

*a. True

b. False

26. Groundwater entering sewer collection pipes through cracks and defective pipe joints is termed as

inﬂow

a. True

*b. False

27. A Combined sewer system is the one that brings the combined domestic and industrial wastewater

ﬂow to the treatment plant

a. True

*b. False

28. Inﬁltration is when groundwater enters the sewage collections systems

*a. True

b. False

29. Deﬁne Inﬁltration and Inﬂow. Discuss the impact of inﬂow and inﬁltration on the wastewater

treatment plant:

30. In wastewater collections what is a gravity system:

31. A sewer system designed to transport only wastewater from homes, industries, institutions and

businesses is called:

a. Combined sewer

*b. Sanitary sewer

c. Service sewer

d. Building sewer

32. Velocity of sewers is usually expressed as:

*a. Feet per second

b. Gallon per minute

c. MGD

d. mg/l

e. sq. ft

33. Which of the following statements is not true regarding a wastewater collection system:

*a. A sewer is designed to allow the waste to ﬂow at a rate of approximately 1 ft/sec.

b. Grease can be serious problem in a collection system.

c. Inﬂow and inﬁltration are frequently problems in older collection systems.

d. High concentrations of hydrogen sulﬁde in a sewer can lead to corrosion of concrete.

e. Scouring can be a problem if wastewater is ﬂowing too fast.

34. Which of the following statements is not true regarding a wastewater collection system:

*a. A sewer is designed to allow the waste to ﬂow at a rate of approximately 1 ft/sec.

b. Grease can be serious problem in a collection system.

c. Inﬂow and inﬁltration are frequently problems in older collection systems.

Chapter 5. Collections

d. High concentrations of hydrogen sulﬁde in a sewer can lead to corrosion of concrete.

e. Scouring can be a problem if wastewater is ﬂowing too fast.

35. A sewer system designed to transport only wastewater from homes, industries, institutions and

businesses is called:

a. Combined sewer

*b. Sanitary sewer

c. Service sewer

d. Building sewer

36. Velocity of sewers is usually expressed as:

*a. Feet per second

b. Gallon per minute

c. MGD

d. mg/l

e. sq. ft

37. Velocity of ﬂow in sewers is usually expressed in terms of

*a. Feet per second

b. Gallon per minute

c. MGD

d. Milligrams per liter

e. Square feet

38. Velocity of ﬂow in sewers is usually expressed in terms of

*a. Feet per second

b. Gallon per minute

c. MGD

d. Milligrams per liter

e. Square feet

39. Inﬁltration is caused by: *a. Cracked pipes b. Improper CCTV operation c. Poor ventilation d. All

of the above

40. Deﬁne inﬁltration & inﬂow, compatible & non-compatible. List 4 major concerns. List 2 correc-

tions.

6. Preliminary Treatment

•

The objective of preliminary treatment is to remove coarse solids and other large materials often

found in raw wastewater

Removal of these materials is necessary to enhance the operation and maintenance of subsequent

treatment units

• Preliminary treatment operations typically include a combination of the following processes:

– Screening

– Grinding or shredding

– Flow measurement

– Grit removal

– Pre-aeration

– Flow equalization

6.1 Process Elements of Preliminary Treatment

6.1.1 Screening

• Screening is typically the ﬁrst unit in a preliminary treatment

•

Screening allows for the capture of coarse solids as pieces of cloths garbage so as to protect pumps

and other units from clogging.

Screens may consist of vertical or inclined bars (bar racks or bar screens), wire mesh or perforated

plates having either circular or rectangular openings.

Screens remove the large, entrained, suspended or ﬂoating solids such as pieces of wood, cloth,

paper, plastics, garbage, etc.

• Debris collected on the screen can be cleaned manually or automatically using chain driven rakes

• The retained material at screens - screenings, is collected and hauled to landﬁll for disposal

•

The quantity of screenings removed varies by location and is a function of the clear opening of the

screen.

Chapter 6. Preliminary Treatment

Barscreen - No rakes

•

Barmuinitors combine the function of a screen and a grinder. The ground material is returned to the

wastewater ﬂow for removal during primary treatment.

6.1.2 Grinding and Shredding

•

Comminutor(Grinder) consist of ﬁxed, rotating or oscillating teeth or blades, acting together to reduce the

solids to a size which will pass through ﬁxed or rotating screens grind rags into small chunks

The comminutors are installed in wastewater channel and they grind the larger solids without actually re-

moving them from the wastewater. These devices may be installed before the screens or as a combination

of screen and cutters (barmunitors).

•

Comminutor

6.1.3 Flow Measurement

Wastewater ﬂow to a treatment plant is not constant but varies in a diurnal (daily) pattern reﬂecting

domestic water use activity.

•

6.1 Process Elements of Preliminary Treatment

Schematic of comminutor placement in a channel

•

Continuous ﬂow measurement is necessary in order to monitor diurnal variations in ﬂow which

may affect treatment plant efﬁciency.

Devices used for ﬂow measurement as part of the preliminary treatment can be placed in a channel

or in a pipe.

Devices for Flow Measurement in Channels

• Weirs:

–

Typically sharp crested weirs which are essentially metal plates installed perpendicular the

ﬂow. The plate may a straight edge, a V-notch, or a trapezoidal opening.

The weir plate in the channel causes an increase in the depth of the water behind the weir.

which is proportional to the ﬂow rate.

The ﬂow rate can be determined by calculation or by reading the corresponding ﬂow value to

the depth of the water, from a chart speciﬁc for that weir.

(a) Weir in a channel

• Parshall Flume:

(b) Weir types

–

Parshall ﬂume uses a narrow section in the channel as the restriction rather than the vertical

plate of a weir.

Chapter 6. Preliminary Treatment

–

Like the weir, the restriction due to the narrow section of the ﬂume, causes an increase in the

depth of the water behind the weir (head) which is proportional to the ﬂow rate.

The ﬂow rate can be determined by calculation or by reading the corresponding ﬂow value to

the depth of the water, from a chart speciﬁc for that Parshall Flume.

(a) Parshall ﬂume

(b) Parshall ﬂume with level sensor

Devices for Flow Measurement in Pipes

• Venturi Tube:

– Measures the difference in pressure in the inlet and center section (throat)

– This pressure difference can then be mathematically converted to a ﬂow rate.

– Works only when a pipe if ﬂowing full

• Magnetic Flow Meter (Magmeter):

–

Magmeter is a pipe spool which has an electromagnetic coil surrounding it. As the wastew-

ater - a conducting material, ﬂows through it, an electrical current is created proportional to

the velocity of the conducting ﬂuid (wastewater).

–

Flow is automatically calculated by multiplying the velocity by the cross- sectional area of

the pipe.

Similar to the venturi meter, magmeter will read accurately only if the magmeter section

of the pipe is ﬂowing full and the wastewater is ﬂowing through it at a certain minimum

velocity.

(a) Magmeter

(b) Piping with magmeters

6.1 Process Elements of Preliminary Treatment

6.1.4 Grit Removal

Grit includes sand, gravel, cinder, eggshells, bone chips, seeds, coffee grounds, and large organic

particles, such as food waste.

•

• Purpose of Grit removal:

– to protect mechanical equipment from abrasion and abnormal wear

– to reduce clogging caused by deposition of grit particles in pipes and channels, and

–

to prevent loading the treatment plant with inert matter that might interfere with the operation

of treatment units such as anaerobic digester and aeration tanks.

• Removal of organic material along with the grit is undesirable for two reasons:

1. It causes odor issues, and

2. Organic matter is a potential source of energy (digester gas)

• Grit Disposal: Grit removed is typically landﬁlled.

• Grit Volume: The volume of grit collected measured in ft³/MG.

• The rate of grit collection can range from 0.5 ft³/MG to 30 ft³/MG.

•

Wastewater plants having a combined collection system must deal with much larger volumes of

grit.

Grit Removal Systems

• HORIZONTAL GRIT CHAMBERS:

–

These are rectangular channels 30 to 60 feet long and the water detention time is between 45

to 90 seconds

–

Water passing through these channels is maintained at a relatively constant velocity of about

1 feet per second (fps) which allows for the grit to settle while keeping the lighter organic

material to stay in suspension and continue on into the primary clariﬁers.

• AERATED GRIT CHAMBERS:

– The 1 fps velocity is maintained by using aerators to create a rolling ﬂow in the tank.

– Aeration is achieved using diffusers located on the bottom of one side of the grit chamber.

–

Aerated grit chambers help create aerobic conditions in septic sewage. Aerobic conditions

help improve the settleability of the sludge and increase both BOD and suspended solids

removal in the primary clariﬁers.

– Much larger and deeper than non-aerated units.

– The detention times are increased to 3 to 5 minutes.

(a) Horizontal grit chamber

(b) Aerated grit chamber

• CYCLONIC/VORTEX GRIT CHAMBER:

– The wastewater ﬂows into a cylinder that tapers to a cone at one end.

Chapter 6. Preliminary Treatment

–

The ﬂow whirls around the inside of the cylinder like a cyclone which causes the heavy grit

to be slinged to the outside and it ultimately settles to the bottom from where it is withdrawn.

(a) Vortex grit chamber design

(b) Vortex grit chamber installed

Grit Removal

–

In the cyclonic/vortex grit chamber, the grit is scoured with water and is removed using

pumps

– For the horizontal and aerated grit systems:

Mechanical augers at the bottom of the grit chamber move the grit to one end of the tank

where grit slurry pumps can pump it out of the tank to a grit separator.

In some cases steep bottom slope is provided which will collect the grit at Central Point

of Removal.

Grit Removal is achieved by air pumps for small aerated grit chambers.

Grit can also be removed by tubular conveyors, buckets type collectors, elevators screws

conveyors, grit pumps and clam shell buckets

6.1.5 Flow Control

•

Flow control is critical for grit removal, speciﬁcally for the horizontal and aerated grit chambers

as excessive or inadequate velocities would lead to poor grit removal or cause excessive organic

material settling along with the grit, respectively

•

As the wastewater ﬂows vary diurnally, it is important that velocity of the wastewater in the grit

chamber should be maintained nearly constant - near 1 fps.

Constant velocity in a grit chamber is achieved by providing a proportional (Sutro) weir at the outlet

end of grit chamber.

The shape of the opening between the plates of a proportional weir is made in such a way that the

discharge is directly proportional to liquid depth in grit chamber resulting in maintaining a constant

velocity of water even a the ﬂow changes.

6.1.6 Pre-aeration

Pre-aeration of the wastewater as part of the preliminary treatment may be provided as a separate

process or increased detention time in an aerated grit chamber.

•

• Pre-aeration provides the follwoing beneﬁts:

– freshens up wastewater by dissolving oxygen thereby reducing the wastewater septicity

–

reduction of septicity allows for better settling - solids and BOD removal, in the following

primary treatment process

6.1 Process Elements of Preliminary Treatment

Diurnal wastewater ﬂow proﬁle

(a) Proportional weir design

(b) Installed Proportional weir

– promotes grease separation which facilitates its removal during primary treatment

6.1.7 Flow Equalization

• Flow equalization involves storing a portion of peak ﬂows for release during low-ﬂow periods

•

It prevents surges and allows for the operation of processes at design ﬂows thus allowing for opti-

mal physical, biological and chemical processes to take place.

•

It results in saving capital costs as the processes may be built with a treatment capacity which is

less than the peak ﬂows

Chapter Assessment

1. A weir can be also be used for measuring ﬂows

*a. True

b. False

2. The process of pre-aeration in no way inﬂuences the degree of settling in a primary clariﬁer

a. True

*b. False

3. Septic sludge has a low pH

*a. True

b. False

4. A Parshall ﬂume measures the velocity of the inﬂuent ﬂow

a. True

*b. False

5. A barminutor frequently operates automatically.

*a. True

b. False

6. A grit chamber with a faster ﬂow velocity than recommended may allow appreciable organic matter

to collect in the grit.

a. True

*b. False

7. A grit chamber with a slower ﬂow velocity than recommended may allow appreciable organics to

Chapter 6. Preliminary Treatment

settle out with the grit.

*a. True

b. False

8. A Parshall Flume is a device used to divide the incoming ﬂow for equal distribution to a plant

having more than one primary clariﬁer.

a. True

*b. False

9. A Parshall ﬂume measures the velocity of the inﬂuent ﬂow

a. True

*b. False

10. A properly operated grit chamber will normally increase the solids loading to the primary clariﬁer

*a. True

b. False

11. A properly operating grit chamber should yield grit that is high in ﬁxed solids (inorganic material)

and low in volatile solids.

*a. True

b. False

12. At most treatment plants preliminary treatment is used to protect pumping equipment and to facili-

tate subsequent treatment processes.

*a. True

b. False

13. A Venturi meter measures the amount of electricity used and should be read when there is a high

electrical demand.

a. True

*b. False

14. A weir can be used as a ﬂow measuring device

*a. True

b. False

15. Hydrogen sulﬁde gas in a moist atmosphere can result in corrosion of concrete structures

*a. True

b. False

16. Inefﬁcient grit removal would tend to cause a decrease in percent volatile solids in raw primary

sludge

*a. True

b. False

17. pH value less than 7 indicates an alkaline or basic condition

a. True

6.1 Process Elements of Preliminary Treatment

*b. False

18. Poor grit removal would affect all of the following: pumps and other mechanical equipment, anaer-

obic digestion, percent volatile and percent total solids in the raw sludge.

*a. True

b. False

19. Pre-aeration improves settleability in primary clariﬁer

*a. True

b. False

20. Pre-aeration of raw wastewater may cause better solids separation and removals

in the primary clariﬁer.

*a. True

b. False

21. Presence of hydrogen sulﬁde cannot always be detected by its characteristic odor

*a. True

b. False

22. In a typical treatment facility, it is necessary to have ﬂow meters on the inﬂuent and efﬂuent to

detect the loss in plant ﬂow.

a. True

*b. False

23. A Venturi meter is a reliable device for measuring ﬂows of either treated or untreated wastewater.

*a. True

b. False

24. A ﬂow measurement device such as a constant differential meter, or the more common name,

rotameter, is used for the measurement of liquids or gases.

*a. True

b. False

25. It is often necessary to pre-chlorinate wastewater to prevent odors. When this is done, it is not

necessary to satisfy a chlorine demand or expect a chlorine residual.

*a. True

b. False

26. Inefﬁcient grit removal would tend to cause an increase in the percent volatile solids in raw sludge.

a. True

*b. False

27. Grit is composed mostly of inorganic material and organic material that is not easily biodegradable

*a. True

b. False

28. A weir can be also be used for measuring ﬂows

Chapter 6. Preliminary Treatment

*a. True

b. False

29. Nderlinehspace1cm is used for controlling the ﬂow velocity in a horizontal grit channel

Correct Answer(s):

a. True

b. False

30. Septic sludge has a low pH

*a. True

b. False

31. Barminutors and comminutors are devices which cut up or shred material which is normally found

in raw wastewater.

*a. True

b. False

32. Barmunitor and comminutor are devices used for cutting up or shredding material normally present

in raw wastewater

*a. True

b. False

33. Coliform testing is typically conducted on a grab sample

a. True

*b. False

34. Comminutors cut up or shred large objects normally found In raw wastewater and remove them

from the wastewater ﬂow.

a. True

*b. False

35. Conductivity is a very useful test for assessing sea water intrusion into sewer lines

*a. True

b. False

36. Fresh wastewater is characterized by a blackish color, foul and unpleasant odors with ﬂoating

materials and suspended solids.

a. True

*b. False

37. Hydrogen sulﬁde in addition to creating an odor nuisance can be an explosion hazard when mixed

with air in certain concentrations.

*a. True

b. False

38. Inﬂow is storm water entering into the sewer system

6.1 Process Elements of Preliminary Treatment

*a. True

b. False

39. Percent efﬁciency of total solids or BOD removal is calculated using the following formula: (In-

Out*100)/(In-(In*Out))

a. True

*b. False

40. Poor grit removal would affect anaerobic digester operation

*a. True

b. False

41. Pre-chlorination is frequently used to disinfect raw wastewater.

a. True

*b. False

@Prechlorination is primarily for reducing septicity

42. Solids removed from preliminary treatment are typically treated in an anaerobic digester

a. True

*b. False

43. The function of a comminutor is to shred rags, paper, wood, and other large wastewater solids and

remove the from the ﬂow.

a. True

*b. False

44. The laboratory measurement of volatile solids is a fair approximation of the organic content of the

wastewater.

*a. True

b. False

45. The size and nature of solids in the wastewater is of no signiﬁcant concern to the wastewater treat-

ment plant operator.

a. True

*b. False

46. The velocity of wastewater ﬂowing through a long channel type of grit chamber may be controlled

by a proportional weir.

*a. True

b. False

47. Total solids are made up of dissolved and suspended solids both of which contain organic and

inorganic matter

*a. True

b. False

48. Typical domestic wastewater BOD content is about 2000mg/l

Chapter 6. Preliminary Treatment

a. True

*b. False

49. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment

a. True

*b. False

50.

51.

52.

matter in wastewater, is normally composed of grit, sand and silt.

a. Colloidal

*b. Inorganic

c. Organic

d. Volatile

is used for controlling the ﬂow velocity in a horizontal grit channel

a. Magmeter

*b. Proportional weir

c. Parshall ﬂume

d. V-notch weir

is used in the wastewater treatment plant to remove debris such as large rocks, branches,

pieces of lumber, leaves, paper, tree roots, etc. from the inﬂuent.

a. Comminutor

*b. Bar Screen

c. Belt ﬁlter

d. Grit chamber

53. A pH probe:

a. Can be used to measure ORP in chlorine disinfection.

b. Is often used to measure hydrogen production in wet wells.

c. Measures, in millivolts, the difference between oxidants like chlorine and reductants such as

organic matter.

*d. Measures hydrogen ion activity in wastewater.

e. Sends a 4-20 mA signal directly to a chlorine controller.

54. A sewer system designed to transport only wastewater from homes, industries, institutions and

businesses is called:

a. Combined sewer

*b. Sanitary sewer

c. Service sewer

d. Building sewer

55. Carryover of grit from the grit chamber may indicate the need to:

*a. Increase rate of settled grit removal from the grit chamber..

b. Decrease the operational depth of the channel.

c. Increase the ﬂow to the primary clariﬁer.

d. Increase the air input to an aerated grit chamber.

6.1 Process Elements of Preliminary Treatment

56. Characteristics that should be measured immediately after the sample is collected are:

a. Velocity and dissolved solids

*b. Temperature, pH and DO

c. TSS and BOD

d. Hardness and alkalinity

57. Flow proportionate composite samples are collected because:

a. The waste characteristics are continually changing

b. The ﬂow is continually changing

*c. The ﬂow and waste characteristics are continually changing

d. This requires less time than grab samples

e. All of the above

58. Grab samples are considered to be representative of the

a. Average daily condition at the sample location

b. Average daily condition in the system

c. System conditions for the two hours before and after the sample was taken

*d. System condition at the time of the sample

59. Grit is composed mostly of which of the following substances?

a. Grease

b. Colloidal solids

c. Rubber goods

*d. Inorganics

e. Plastics

60. Organisms in wastewater that are not harmful to humans but are indicators of diseases are:

a. Pathogens

b. Viruses

*c. Coliform

d. Bacteria

61. Which of the following pollutants would be removed to the extent in an efﬁciently operating grit

chamber?

a. Egg shells

b. Seeds

c. Oils and grease

*d. Sand

e. Fixed solids

62. Proportional weirs usually are located at:

a. Immediately after the barscreens

b. Primary clariﬁers

c. Aerobic digester scum boxes

*d. Grit chambers

e. Inside the Parshall ﬂume

Chapter 6. Preliminary Treatment

63. Which of the following process units is not usually considered to be a preliminary treatment unit?

a. Grit chamber

b. Bar screen

c. Comminutor

d. Bar rack

*e. A meniscus

64. Which of the following would be included in the pretreatment unit?

a. preaeration

b. grit removal

c. screening

d. comminutor

*e. all of the above

65. As grit accumulates in a recetangular channel or chamber, the velocity of the inﬂuent wastewater:

*a. increases

b. decreases

c. remains constant

d. behaves independent of inﬂow

e. none of the above

66. Grit usually contains some organic matter which decomposes and creates odors. To facilitate

disposal without nuisance, the organic matter is removed by washing methods. Commonly used is:

a. aeration

b. elutriation

c. vacuum ﬁltration

d. wet oxidation

*e. none of the above

67. The ﬂow through rate for grit channels is usually:

a. 20 seconds to 1.0 minute

b. 2 feet per second

*c. 1 foot per second

d. 30 days, depending on temperature

e. none of the above

68. Given the following data, what is the most likely cause of the mechanically cleaned bar-screen

problem?

DATA: Normal dry weather ﬂow

Motor running but unit not operating

Drive chain excessively tight

Water differential across screen above 6 inches

Controls on automatic.

Pin sheared or automatic clutch tripped.

a. Bubble tube malfunction

b. Flow too high

6.1 Process Elements of Preliminary Treatment

*c. Rocks lodged in screen

d. Suction channel level too high

69. In which of the following types of wastewater would settling occur most rapidly?

*a. Cold wastewater

b. Old (stale) wastewater

c. Septic wastewater

d. Strong wastewater

70. Given the following data, whatis the most likely cause of

the grit separator problem?

DATA: ·Lower than normal ﬂow of water and grit from apex

High separation chamber pressure

*a. Grit pump suction clogged.

b. Partial obstruction near apex

c. Stick lodged in separation chamber

d. Vortex ﬁnder worn

71. What is the most likely cause of an aerated grit chamber problem if the grit pump is in automatic

mode and running, the suction valve is wide open, the pressure on discharge line is low and erratic,

and less than normal water and grit are discharging from discharge line?

a. Discharge check valve partially plugged.

b. Grit classiﬁer partially plugged.

*c. Grit pump suction line partially plugged.

d. Malfunctioning air supply to grit chamber causing pump to get air bound.

72. Usually, preliminary treatment includes removal of most of the:

a. Pathogenic bacteria.

b. Biodegradable organics.

c. Settleable solids.

d. Dissolved solids.

*e. Grit.

73. A spray nozzle on the mechanically cleaned screen has become plugged. To ensure your safety,

prior to entering the screen housing to repair the nozzle, you should

a. Leave note on breaker panel of repair being made

b. Request assistance for repair •

*c. Turn off and lock motor control

d. Turn off local control switch

74. An aerobic treatment process is one that requires the presence of:

a. Ozone

b. organic oxygen

c. no oxygen

d. combined oxygen

*e. dissolved oxygen

100

Chapter 6. Preliminary Treatment

75. Which of the following should not normally be a signiﬁcant part of grit

a. Sand.

b. Rocks

c. Eggshells.

*d. Fecal Matter.

76. Samples collected over several hours during the day and combined are known as:

*a. Composite samples.

b. Grab samples.

c. Deep samples.

d. Periodic samples.

77. Which of the following is not a characteristic of hydrogen sulﬁde?

a. Foul odors

*b. Lighter than air

c. Toxic

d. Corrosiveness

e. Explosiveness

78. The following device is used to measure the ﬂow of wastewater:

a. Comminutor.

b. Comparator.

*c. Parshall ﬂume.

d. Sluice gate.

79. A device called an Imhoff cone is commonly used to measure settleable solids in:

a. Percent

*b. mL/L

c. mg/L

d. ppm

e. SVI units

80. The proper operation of an aerated grit removal process will:

a. Cause material with a speciﬁc gravity of greater than 1.0 to settle.

b. Cause sand and other non-organics to settle and keep organic material in suspension.

c. Help to freshen stale or septic wastewater.

*d. b and c

81. Velocity of sewers is usually expressed as:

*a. Feet per second

b. Gallon per minute

c. MGD

d. mg/l

e. sq. ft

82. Which of the following statements is not true regarding a wastewater collection system:

6.1 Process Elements of Preliminary Treatment

101

*a. A sewer is designed to allow the waste to ﬂow at a rate of approximately 1 ft/sec.

b. Grease can be serious problem in a collection system.

c. Inﬂow and inﬁltration are frequently problems in older collection systems.

d. High concentrations of hydrogen sulﬁde in a sewer can lead to corrosion of concrete.

e. Scouring can be a problem if wastewater is ﬂowing too fast.

83. In order for the grit to settle in a non-aerated grit chamber, the average velocity should be kept near:

*a. 1 ft/sec .

b. 2 ft/sec.

c. 5 ft/sec.

d. 1 ft/min.

e. 2 ft/min.

7. Primary Treatment

• Synonyms: primary treatment basin, primary clariﬁer, sedimentation basin, primaries, clariﬁer

• Primary treatment is after preliminary treatment and before secondary treatment

• Its two main objectives are:

– Remove settleable solids

– Remove ﬂoatable solids

•

This is a physical process which relies on the physical properties - how heavy or light the sus-

pended solids particles are to effect its separation

Provides quiescent conditions for the inﬂuent wastewater for the heavier solids to settle and the

lighter solids to ﬂoat

• Removes settleable solids and ﬂoatables

• Settled solids are removed as sludge from the bottom of the clariﬁer

• Floatable solids including oil and grease are also removed, as scum from the surface

• The shape of the primary clariﬁer is either rectangular or circular

•

Effective solids removal in the primary clariﬁers will reduce the loading on the expensive sec-

ondary treatment process.

•

The amount of solids removed during primary treatment may be enhanced by chemical addition

- ferric or ferrous chloride as a coagulant and anionic polymer as the ﬂocculant. This is called

Chemically Enhanced Primary Treatment (CEPT).

• Typical Removal Rates:

–

BOD removal – 25% to 40% and about 60% with CEPT

Suspended solids (SS) removal – 40% to 60% and about 75% with CEPT

Settleable Solids removal - >90%

104

Chapter 7. Primary Treatment

Cross section of a Rectangular Clariﬁer

Schematic cross section of a circular clariﬁer

Cross section of a circular clariﬁer

7.1 Clariﬁer Zones

105

7.1 Clariﬁer Zones

7.1.1 Inlet Zone

•

Inlet Zone is where the water enters the end of a rectangular tank, or the center of a circular or

square tank.

• The Inlet Zone is designed to accomplish two objectives:

1. Reduce the velocity (dissipate energy in the incoming water)

2. Distribute the ﬂow evenly

•

The inlet zone is equipped with a bafﬂe. Inlet bafﬂe reduces the velocity of the inﬂuent ﬂow, pre-

vent short circuiting which could cause solids being carried over to secondary treatment.

–

Circular tanks are equipped with a collar-type circular bafﬂe that directs the water down as it

enters the center of the tank.

–

Rectangular tanks will have a plate bafﬂe in front of the opening for the wastewater ﬂow into

the clariﬁer and another bafﬂe just upstream - a perforated wall or a picket fence type bafﬂe

that spreads the water laterally across the inlet end of the tank.

(a) Rectangular clariﬁer inﬂuent bafﬂe

(b) Circular clariﬁer inﬂuent bafﬂe

7.1.2 Settling Zone

• This is the largest portion of the tank where solids settle.

• The water velocity is reduced to 0.03-0.05 fps and the detention time is about 1.5 to 2 hours.

•

A clariﬁer is said to be short circuiting if the velocity of the water is greater in some sections than

in others. The water passing through the higher velocity region will have a reduced detention time

and settleable solids will carry through with this water as it goes over the weir. Short circuiting is

prevented by appropriately designing inlet bafﬂes and weir plates (at the Outlet Zone).

7.1.3 Sludge Zone

• Sludge zone is the bottom of the tank where the settled sludge collects and compacts.

•

Sludge blanket depth should be measured and sludge should be removed at least every shift. A

desirable blanket depth is typically established and the sludge pumping rate and regimen is estab-

lished to maintain that desired sludge blanket level.

• Sludge rakes push the sludge to one end or the center of the tank so that it can be pumped out.

•

The rake drive is usually equipped with a torque indicator. A shear pin in the drive shaft will break

to prevent damage to the gearbox or drive shaft.

106

Chapter 7. Primary Treatment

•

Failure to remove sludge often enough will result in the sludge becoming septic releasing gas

bubbles which hinders the sludge settling and also result in causing odor problems.

The sludge from the primary clariﬁers needs to be stabilized prior to its disposal. The sludge

(solids) from the primary clariﬁers are mixed with the solids from the secondary treatment pro-

cess and stabilized typically using a sludge digestion process.

7.1.4 Skimming Zone

• The skimming zone is at the surface of the tank for scum removal

• Lighter solids and greases ﬂoat to the surface of the clariﬁer as scum

•

In Circular Clariﬁers: Floating matter is skimmed by a skimmer arm that is supported by the sludge

rake and rotates with it around the tank. The ﬂoating matter is pushed over the beach plate by the

wipers attached to the skimmer arm and into a scum box attached to the tank wall.

In Rectangular Clariﬁers: The ﬂights act as skimmers when the chain brings them to the surface

and pushes the scum towards the scum troughs. The scum trough may be designed to rotate (tip)

periodically for the scum to ﬂow in from the water surface.

The scum collected from the primary clariﬁers is sent to the digester for treatment along with the

sludge removed.

Rectangular clariﬁer scum trough

•

The ﬂights in the rectangular clariﬁer are supported at the top by two parallel rails running along

the length of the clariﬁer.

There are wear plates (strips) installed at the clarifer bottom and on top of the rails to prevent the

ﬂights from riding directly on those surfaces. To reduce friction, the ﬂights have a wearing shoe

7.1 Clariﬁer Zones

107

attached.

• Both the wear strip and the wearing shoe are disposable items and are replaced at ﬁxed intervals.

Rectangular clariﬁer ﬂights

7.1.5 Outlet Zone

•

This is the part of the clariﬁer where the settled water leaves to go to the secondary treatment

processes.

A channel called the efﬂuent launder collects the efﬂuent ﬂow and directs it to the primary efﬂuent

piping.

Weirs are installed along the edge of the efﬂuent launder channel to skim the water evenly off the

surface of the tank. The most common type of efﬂuent weir is a V-notch (or saw-tooth) weir. A

V- notch weir is a plate that has notches, about 2-3 inches deep, cut in it every 6-8 inches. If the

weir is clean and level, it will remove water evenly all the way around the edge of the tank. This

minimizes the upward velocities near the efﬂuent launder and improves removal efﬁciencies. If

the weir plate is not level or part of the weir becomes clogged with slime or debris, short-circuiting

will result because more water will pass over the low side or the clean notches of the weir. Short-

circuiting will cause poor settling and uneven sludge blanket buildup.

• In rectangular tanks the water leaves at the end opposite the inﬂuent.

• In circular tanks the water leaves at the edge of the tank.

•

Also, in the circular clariﬁers, an efﬂuent bafﬂe, just upstream of the weir, is installed to prevent

ﬂoating solids from going over the weir.

108

Chapter 7. Primary Treatment

Circular clariﬁer skimmer arm, efﬂuent bafﬂe and v-notch weir

Rectangular clariﬁer v-notch weir and launder

7.2 Sludge Pumping

109

7.2 Sludge Pumping

•

The sludge pumping from the clariﬁer must be adequate to prevent sludge from going septic. Septic

sludges are much more difﬁcult to thicken or de-water and cause odor issues.

• Primary sludge normally averages 4-6% solids. Generally positive displacement pumps are used for

primary sludge

• The pumping cycles must be designed to provide the thickest sludge possible.

•

Excessive pumping or pumping without building solids to build up leads to pumping thinner (more

water) sludge.

7.3 Design Parameters

• Clariﬁer depth – 8 to 12 feet

• Hydraulic or surface loading

– This rate is important to ensure good settleable solids removal efﬁciency

– It is expressed in terms of gallons per day per square foot (gpd/sq ft) of tank surface area

–

Typical surface loading rate used for the design of primary clariﬁers range between 300 to

1,400 gpd/sq ft, depending on the nature of the solids and the treatment requirements. Lower

loading rates are frequently used in small plants in cold climates. In warm regions, low rates

may cause excessive detention which could lead to septicity.

7.4 Advance primary treatment (APT)

Synonyms: Advance primary treatment (APT), Chemically enhanced primary treatment (CEPT), Physical-

chemical treatment (Phys-chem)

7.4.1 Background

•

Suspended solids present in wastewater are typically coated with bacterial slime and biological

metabolic products which are negatively charged. A signiﬁcant portion of the suspended solids in

wastewater do not settle easily due to gravity as:

1. the biological mass and the associated byproduct gases produced makes these particles buoy-

ant,

2. the negative electrostatic charges on these particles cause these particles to be in constant

state of motion due to electrostatic repulsion

•

Advance primary treatment (APT) also known as Chemically enhanced primary treatment (CEPT)

or Physical-chemical treatment (Phys-chem), involves chemical addition to the primary inﬂuent

ﬂow to enhance primary treatment TSS and BOD removal efﬁciencies

a normal primary treatment process typically removes 40 to 60% TSS and 25 to 40% BOD. TSS

and BOD removal efﬁciencies of over 80% and 60% respectively, may be achieved by the use of

APT.

additional cost incurred for the chemical addition is in most cases is offsetted by the beneﬁts which

include:

1. by removing more BOD in the primary treatment cost associated with secondary treatment is

reduced

2. primary BOD is more easy to digest than the secondary biomass thus the digester gas pro-

duction is increased and digested solids production is lowered thus saving biosolids hauling

110

Chapter 7. Primary Treatment

cost

3. residual ferric chloride in the primary sludge provides H₂S and struvite control in the solids

treatment processes

7.4.2 Process/mechanism

APT is a two step chemical process:

Coagulation

•

Coagulation is the process by which the negative charge on these particles is reduced lowering the

repulsion forces, by the use of a chemical such as ferric chloride and alum.

The concentration of the coagulant required is dependent on the strength of the wastewater and the

conveyance time of the wastewater.

Typically the coagulant is added immediately after the grit chambers so the conveyance from the

grit chambers to the primary clariﬁer provides adequate contact time and mixing.

• Parameters to ensure optimal coagulation are:

– Appropriate coagulation concentration

– Adequate mixing energy, and

– Adequate contact time

• Overdosing the coagulant will adversely effect the settleability.

• Typical ferric chloride dosage for coagulation range from 12 to 22 mg/l.

Flocculation

•

Flocculation uses an anionic polymer - polymer which has negatively charged groups, to bridge the

coagulated particles to a size which will settle in the primary clariﬁer.

The ﬂocculated particles are prone to shearing thus the polymer is gently folded in with the coagu-

lated wastewater just prior to entry into the primary clariﬁer

CEPT Schematic

Untreated Primary Inluent

Coagulation

Flocculation

Chapter Assessment

1. An Imhoff cone is often used to measure the effectiveness of primary sedimentation.

a. True

*b. False

2. An inadequate detention time in a primary clariﬁer would result in increased solids pumping to

digesters

a. True

*b. False

3. A well-operated primary clariﬁer, will remove between 90 and 95 percent of the inﬂuent settleable

solids.

*a. True

b. False

4. Bafﬂes ahead of outlet weirs in primary tanks serve as a physical barrier to inhibit the loss of

ﬂoating solids and grease.

*a. True

b. False

5. Bafﬂes upstream of outlet weirs in primary tanks serve as a physical barrier to inhibit the loss of

ﬂoating solids and grease.

*a. True

b. False

6. Detention time is the time required to ﬁll a tank at a given ﬂow.

*a. True

b. False

7. Gas bubbles rising to the surface in a primary clariﬁer are normally an indication that the sludge in

the unit is undergoing proper settling

a. True

*b. False

112

8. Hydraulic loading to a clariﬁer is expressed as gallons/day.

Chapter 7. Primary Treatment

a. True

*b. False

9. It is common for primary clariﬁers to remove 80-85% of the BODand90-95% of the TSS.

a. True

*b. False

10. Operators may use sludge blanket levels in primary clariﬁers to adjust sludge pumping rates.

*a. True

b. False

11. The weir overﬂow rate of a primary clariﬁer is expressed as gal/day/ft2 (gallons per day per square

feet)

a. True

*b. False

12. Primary treatment is a process which allows substances that will settle or ﬂoat to be separated from

the wastewater being treated.

*a. True

b. False

13. A well operating primary clariﬁer will generally remove nearly all of the inﬂuent suspened solids.

a. True

*b. False

14. Sludges from primary clariﬁers are usual ly more dense than sludges from secondary clariﬁers.

*a. True

b. False

15. The generally accepted range of BOD removal in a well operating primary clariﬁer would be 40 -

60% .

a. True

*b. False

16. Wooden ﬂights in a rectangular clariﬁer usually are equipped with wearing shoes.

*a. True

b. False

17. The typical percentage of BOD which should be removed in primary clariﬁers is in the range of

about 90%-99%

a. True

*b. False

18. If the collector mechanism on a circular clariﬁer stalls, it would be advisable to reverse the motor

and back up the collector.

a. True

*b. False

19. Raw sludge drawn from a primary sedimentation tank normally would contain 1,000 mg/l - 3,000

mg/1 of settleable solids.

a. True

*b. False

20. When pumping primary clariﬁer scum to a holding tank, there should be no additional problems in

cold weather.

a. True

7.4 Advance primary treatment (APT)

113

*b. False

21. If a shear pin for a particular installation is constantly bending, it would be proper and advisable to

replace it with a pin of greater shear strength.

a. True

*b. False

22. Septic sludge generally has a low pH.

*a. True

b. False

23. Short circuiting in a primary clariﬁer occurs mainly because of the efﬂuent weirs being plugged or

not level

*a. True

b. False

24. Wear shoes are found in circular clariﬁers

a. True

*b. False

25. Hydraulic loading is the number of gallons ﬂowing each day through one cubic foot volume of the

clariﬁer

a. True

*b. False

26. Clariﬁer detention time is the time required to ﬁll the clariﬁer at a given ﬂow rate

*a. True

b. False

27. Main objective of primary treatment is to remove organics present in wastewater

a. True

*b. False

28. Raw sludge pumped from a primary clariﬁer generally contains about 3% to 6% total solids.

*a. True

b. False

29. The main function of launders in a rectangular sedimentation tank is to prevent scum and other

ﬂoatables from leaving with the primary efﬂuent.

*a. True

b. False

30. The solids in primary sludge contain both volatile and ﬁxed parts.

*a. True

b. False

31. Uneven weirs (heights) may result In short circuiting in a primary clariﬁer

*a. True

b. False

32. “V” -notch weirs attached to a primary efﬂuent launder measure the ﬂow rate.

*a. True

b. False

33. A well-operated primary clariﬁer, will remove between 90 and 95 percent of the inﬂuent settleable

solids.

*a. True

b. False

114

Chapter 7. Primary Treatment

34. A well operating primary clariﬁer will generally remove nearly all of the inﬂuent settleable solids.

*a. True

b. False

35. In a well-designed, efﬁciently operated wastewater treatment facility, most of the colloidal and

dissolved matter in wastewater will be removed in the primary clariﬁers

a. True

*b. False

36. In some primary clariﬁers water or aIr sprays may be used to push scum toward the scum removal

point.

*a. True

b. False

37. The main purpose of primary treatment is to remove the materials that might damage downstream

equipment and/or the pumps.

a. True

*b. False

38. The volume of sludge to be removed from a primary settling tank maybe estimated by measuring

the primary inﬂuent and efﬂuent settleable solids and wastewater ﬂow.

*a. True

b. False

39. A device called an Imhoff cone is commonly used to measure settleable solids in:

a. Percent

*b. mL/L

c. mg/L

d. ppm

e. SVI units

40. An efﬁcient primary clariﬁer is expected to remove what percent of the inﬂuent settleable solids?

a. 2 to 6 %

b. 20 to 40 %

c. 40 to 50 %

d. 60 to 75 %

*e. 90 % or better

41. A primary clariﬁer does not have adequate detention time. Which of the following would result?

a. Decreased organic leading on secondary unit

b. Overloading of collector or ﬂight drive motor

c. Increased solids pumped to digester

*d. Low BOD removal

e. Reduced suspended solids in aeration tank.

42. If the sludge depth in a secondary sedimentation tank is too high, what will happen?

a. Decreased turbidity in efﬂuent.

b. Return activated sludge will have lower oxygen demand

c. Settleable solids from aeration tank will increase

*d. Sludge may become septic

43. Odors associated with septic sludge are usually caused by

a. A neutral pH

b. Inorganic matter

7.4 Advance primary treatment (APT)

115

c. Short retention time in collection system

d. Active aerobic bacteria

*e. Active anaerobic bacteria

44. The main objectives of primary sedimentation are to remove:

a. Finely divided particles and dissolved organics

b. TDS and colloidal solids

c. BOD, COD, and SS

*d. Settleable solids and ﬂoatables

e. Detritus and volatile organics

45. The withdrawal of sludge from a primary clariﬁer should be slow in order to:

a. conserve electricity

*b. prevent pulling too much water with the sludge

c. keep the BOD stabilized

d. not disturb the bacteria in the clariﬁer

e. protect the pump

46. The withdrawal of sludge from a clariﬁer should be slow in order to:

a. protect the pump.

b. conserve electricity.

*c. prevent the pulling of too much water with the sludge.

d. not disturb the bacteria in the digester.

e. avoid breaking the water seal in the digester.

47. Which of the following terms refers to a hydraulic condition, typically indicated by billowing solids

ﬂowing over the efﬂuent weir, where a portion of the ﬂow through a clariﬁer experiences a much

shorter detention time than the rest of the wastewater in the tank?

a. surging

*b. shortcircuiting

c. overload

d. dispersion

48. When wastewater enters a rectangular clariﬁer, it is often evenly dispersed across the entire width

of the tank by means of:

*a. a bafﬂe.

b. a proportional weir.

c. a submerged launder.

d. a dome diffuser.

e. a hydraulic pump.

49. Rectangular primary clariﬁers have wooden or plastic ﬂights, which are equipped with:

a. shear pins.

b. friction skirts.

*c. wear shoes.

d. grease nipples.

e. sludge hinges.

50. The main function of an inlet bafﬂe in a settling tank is to:

*a. Reduce velocity and disperse the ﬂow

b. Increase velocity to prevent excessive settling near the inlet

c. Remove scum from the wastewater

116

Chapter 7. Primary Treatment

d. Protect the scrapping mechanism from damage by excessive velocities.

51. If short circuiting occurs in a clariﬁer, the operator should

*a. Identify the cause

b. Change fuses.

c. Increase the sludge drawoff

d. Restart the pump.

52. The expected range of BOD removal in a well operated primary clariﬁer is:

a. 10 to 20%

*b. 25 to 40%

c. 40 to 60%

d. 60 to 80%

53. A treatment plant process using sedimentation after screening and grit removal and before aerobic

treatment is known as:

a. preliminary treatment.

*b. primary treatment.

c. secondary treatment.

d. RBC treatment.

e. advanced wastewater treatment.

54. Sludge gasiﬁcation in a primary clariﬁer may be the result of:

a. hydraulic overloading

b. low inﬂuent BOD concentrations

*c. infrequent pumping of sludge

d. too short of detention time

55. The withdrawal of sludge from a primary clariﬁer should be slow in order to:

a. conserve electricity

*b. prevent pulling too much water with the sludge

c. keep the BOD stabilized

d. not disturb the bacteria in the clariﬁer

e. protect the pump

56. Primary sedimentation will remove most of the:

a. BOD and suspended solids.

b. settleable solids and suspended solids.

c. grit and raw sludge.

*d. ﬂoatable and settleable solids.

e. suspended solids and pathogens.

57. The time it takes for a unit volume of wastewater to pass entirely through a primary clariﬁer is

called:

*a. detention time

b. hydraulic loading rate

c. overﬂow time

d. weir loading rate

58. Primary treatment units are designed to remove settleable solids from the wastewater stream by:

a. biological treatment

b. chemical addition

c. bioﬁltration

7.4 Advance primary treatment (APT)

117

*d. gravity sedimentation

e. comminuting devices

59. Which one of the following factors would have least inﬂuence on the settleability of solids in a

clariﬁer:

a. detention time

b. ﬂow velocity

c. short-circuiting

d. temperature

*e. soluble BOD

60. The least acceptable method of handling scum and skimmings from a primary clariﬁer is:

a. burning

b. burial

c. pumping to a digester

*d. discharging into the efﬂuent

e. hauling to a sanitary landﬁll

61. Given the following data, what is the most likely cause of the primary sedimentation tank problem?

DATA: Raw sludge pumps run 10 minutes in each hour Raw sludge has 3% total solids at start

of pumping cycle, 2% total solids at end. No sludge accumulation on tank ﬂoor Slight sludge

accumulation in sludge hopper

a. Raw sludge pumping duration too short

*b. Raw sludge pumping duration too long

c. Raw sludge total solids too high

d. Sludge collectors running too long

62. Raw sludge pumping cycles that are too short will

*a. Cause wastestream going to secondary treatment to turn dark.

b. Decrease return and waste sludge suspended solids.

c. Increase amount of grit in raw sludge.

d. Increase rags in raw sludge.

63. Raw sludge should be removed from a primary settling tank at any plant

a. At least hourly.

b. Not more often than once a week.

*c. At least once a day.

d. Whenever sludge rises to the surface.

118

Chapter 7. Primary Treatment

64. You are the operations supervisor at a modern 27 MGD (average dry weather ﬂow) conventional

activated sludge wastewater treatment plant. During a recent plant expansion, four (4) new primary

sedimentation tanks (160 ft. long x 25 ft. wide x 12 ft. deep) were built. These tanks have been

tested and are ready for service. These new tanks are to be put on line while six old sedimentation

tanks are to be taken out of service so that they may be inspected and repaired. The primary inﬂu-

ent channel and other mechanical equipment (e.g. pumps, collector mechanisms, ﬂights, etc.) will

also be inspected and repaired. It is estimated that the total down time will be one month or more.

Because of the design of the primary inﬂuent channel, all six of the old sedimentation tanks must

be taken out of service at one time.

The table below summarizes the calculated surface loading rates and detention times with the

original and when only four tanks are on line.

FLOW CONDITION

Ave. Daily Flow

Peak Flow

SURFACE LOADING

1688 gpd/ ft²

DETENTION TIME

1.3 hours

2531 gpd/ ft²

0.85 hours

Note: Peak ﬂow is 1.5 times the average daily ﬂow.

Write a memo to your shift supervisors in which you identify and brieﬂy discuss FIVE (5) signiﬁ-

cant concerns in regard to a plan of action (POA) for removal and inspection of the old sedimenta-

tion tanks. State all assumptions.

7.4 Advance primary treatment (APT)

119

Response:

Assessing the clariﬁer surface loading rate for the new clariﬁers:

Surface area of new tanks= 160ft ∗25 ft = 4,000ft²/tank = 16,000ft²total surface area - four

new clariﬁers

Clariﬁer surface loading - average dry ﬂow condition:

ꢀ

ꢁ

Clari fier influent flow(gpd)

Clari fier sur face area(ft )

gpd

27∗10 gpd

gpd

=⇒

= 1,688

16,000ft

Assessing the clariﬁer detention time for the new clariﬁers:

Volume of new tanks= 160 ft ∗ 25 ft ∗ 12 ft = 48,000 ft³/tank = 288,000ft³total volume - four

new clariﬁers

Clariﬁer detention time - average dry ﬂow condition:

Clari fier volume(ft )

288,000ft

(hrs) =

=⇒

= 1.91 hrs

gal

day

flow(

)

27∗10

∗

day 24hrs 7.48gal

Both, detention time & the surface loading rate of the new clariﬁers is comparable to the existing

ones - so no concern from that perspective.

i Ensure the efﬂuent weirs in the new clariﬁers are level to ensure there is no short circuiting

ii Ensure that the new clariﬁer mechanical equipment and its associated control systems are

tested to ensure reliable operations prior to the switch.

iii For the inspection/repair of the old tanks, ensure lockout-tagout and proper conﬁned entry

procedures are followed

120

Chapter 7. Primary Treatment

65. Chemical enhanced primary treatment (CEPT) is used at a number of conventional activated sludge

planes. CEPT, also called advanced primary treatment, typically involves the application of liquid

ferric chloride and an anion polymer ahead of the primary sedimentation tank. Although the long-

term addition of chemicals to a wastewater ﬂow can be costly, the beneﬁts of CEPT may outweigh

its cost.

Answer the fol- lowing question about advanced primary treatment:

(a) Identify two operational considerations in order optimize CEPT.

(b) Identify and explain two ways in which CEPT could result in energy saving in a convention

activated sludge plant that uses cogeneration

addition of ferric chloride. Brieﬂy explain how ferric chloride is able to yield these beneﬁts.

Response:

a Identify two operational considerations in order optimize CEPT.

• Optimal ferric chloride and polymer dosage

• Adequate mixing time and energy for ensuring proper coagulation by ferric chloride

•

Ensure the polymer is gently folded into the coagulated wastewater just prior to entering

the clariﬁer.

Identify and explain two ways in which CEPT could result in energy saving in a convention

activated sludge plant that uses cogeneration.

•

By increasing BOD removal in the primary clariﬁer, the inﬂuent BOD loading to the

secondary treatment is lower reducing the oxygen requirements thus resulting in energy

savings.

•

The organic material in the sludge feed to the digester has a higher proportion of pri-

mary sludge which is more easily digestable thus producing more digester gas for cogen-

eration.

Identify two additional beneﬁts to anaerobic digester operations that may result from the

addition of ferric chloride. Brieﬂy explain how ferric chloride is able to yield these beneﬁts.

•

The residual ferric chloride in the primary sludge feed allows for the control of H₂S

concentration in the digester gas

•

Ferric chloride also helps in minimizing struvite forming potential in the digester due to

its ability to remove phosphate, one of the key elements of struvite.

7.4 Advance primary treatment (APT)

121

66. Recently your activated sludge plant has acquired a two 2-meter belt press to dewater anaerobically

digester sludge.

(a) List four (4) factors, other than belt speed, which generally affect the belt press performance.

(b) The belt press is running. You observe what one of your operator’s calls washout. Deﬁne

"washout."

(d) It has been said that the "the ideal belt speed is the slowest the operator can maintain without

causing washout." Explain why this is so.

Response:

a List four (4) factors, other than belt speed, which generally affect the belt press performance.

•

Primary sludge dewaters better than secondary sludge. Digested sludge with a higher

proportion of primary will produce drier (higher solids content) dewatered biosolids.

• Optimal dosage and adequate mixing of polymer with the sludge.

• Cleanliness of the belts.

• Adequate and even hydraulic pressure to ensure water is squeezed out from the sludge.

b Identify two possible cause of washout.

• Inadequate polymer dosing.

• Hydraulic overloading.

• Blinded gravity zone belt.

• Slow belt speed

It has been said that the "the ideal belt speed is the slowest the operator can maintain with-

out causing washout." Explain why this is so.

As the belt speed is slowed, the water from the sludge will have more time to drain producing

a drier cake. However, due to the slow belt speed, solids will build up on the belt preventing

the water from draining. Washout occurs when the belt speed is slowed to a point when there

is so much standing water that it ﬂows over the sides of the belt.

8. Introduction to Secondary Treatment

•

While preliminary and primary treatment processes are designed primarily to remove solids from

wastewater, secondary treatment is for the removal of organics.

• Secondary treatment involves:

–

biological conversion of the dissolved and suspended organics in wastewater into biomass,

and

–

physical settling (separation) process where the solids including the biomass formed during

secondary treatment is separated and removed from the treated wastewater.

•

With the removal of gross solids in the preliminary treatment followed by removable of settleable

solids in the primary clariﬁers and the removal of dissolved and suspended organics in the sec-

ondary treatment processes, the wastewater is considered treated.

•

Secondary treated wastewater is typically disposed or treated further for reuse or disposal (depend-

ing upon the end use/application and the NPDES permit stipulations).

The solids (biomass) removed from the secondary treatment is typically mixed with the solids from

primary treatment and stabilized using a solids treatment process like sludge digestion prior to its

disposal.

Secondary treatment process incorporates one of the following three approaches:

8.1 Fixed ﬁlm system

•

Here the microorganisms responsible for the treatment, grow on substrates such as rocks, sand or

plastic.

When the wastewater is spread over the substrate, the microorganisms up-take the organics present

in the wastewater

Example of this secondary treatment process include trickling ﬁlters and rotating biological contac-

124

Chapter 8. Introduction to Secondary Treatment

tors

8.2 Suspended Growth System

•

In this type of secondary treatment, the microbes are suspended in the wastewater ﬂow being

treated.

•

Air or oxygen is supplied to maintain an aerobic environment and to keep the microorganisms in

suspension.

• Example of this secondary treatment approach include the activated sludge treatment process

8.3 Pond System

Similar to the suspended growth, stabilization ponds are large man made bodies of water which treat

wastewater using mainly natural processes including sunlight, algae and microorganisms.

9. Trickling Filters

9.1 Theoretical Background

Trickling ﬁlter is a ﬁxed ﬁlm secondary treatment process wherein the organic content of the wastewater is

removed using biological growth attached to an inert media such as lava rock or plastic

•

In a trickling ﬁlter, the wastewater is sprayed evenly on the surface of the media with a rotary type

distributor with oriﬁces

The wastewater percolates through the media bed, where it comes in contact with biological slime

growth – zoogleal ﬁlm (zooglea)

The aerobic biomass - bacteria, protozoa and other microoorganisms in the zooglea capture and

consume the suspended and dissolved organics from the wastewater.

The microorganisms metabolize the organics and in the process produce more microbial mass

resulting in increasing the thickness of the zoogleal layer.

The thickness of the zoogleal layer can only increase to a point until the wastewater ﬂow – hy-

draulic load, shears the slime layer – “sloughs off” and is carried out as part of the efﬂuent ﬂow as

sloughing.

•

The treated wastewater cascades from the bottom of the media into the underdrain system – lower

portion of the TF comprised of columns which support the media base. The underdrain has a

sloping ﬂoor to direct the cascading water into a center channel .

The clariﬁer allows for the separation (settling) of the of the solids (sloughed off material). The

settled solids is removed - typically pumped to a digester and the clariﬁed efﬂuent ﬂows out of the

clariﬁer.

The source of oxygen to support the aerobic growth is from the oxygen dissolved in the wastewater

as it is sprayed over the media and from the air currents due to the downward ﬂow of the wastew-

ater and the temperature difference between ambient and the interior of the trickling ﬁlter. Forced

ventilation system may be designed as part of the trickling ﬁlter

• Word trickling “ﬁlter” is a misnomer - no ﬁltration is involved

126

Chapter 9. Trickling Filters

• Advantage includes process simplicity and lower costs

• Disadvantage include BOD removal efﬁciency of only about 80-85

•

The media may be rock, slag, coal, bricks, redwood blocks, molded plastic, or any other sound

durable material.

The media depth ranges from about three to eight feet for rock media trickling ﬁlters and 15 to 30

feet for synthetic media.

The media needs to be uniformly sized and have adequate empty spaces (voids) to ensure maintain-

ing aerobic condition necessary for the survival of biomass.

Pre-fabricated (synthetic) media - similar to the one shown below, has an advantage over the

"dumped" type media such as lava rock of providing a greater surface area per volume upon which

the zoologeal ﬁlm may grow while providing ample void space for the free circulation of air.

Sometimes, due to inadequate hydraulic loading, portions of the zoogleal layer may become too

thick and oxygen cannot penetrate its full depth, causing odor issues.

•

9.2 Trickling Filter Recirculation

Recirculation - where a portion of the treated wastewater is returned back as the feed to the TF. The

parameter recirculation ratio is calculated to quantify the recirculation ﬂow. Recirculation ratio is a ratio

of the recirculated ﬂow - Q_Rto the inﬂuent ﬂow Q_I. Recirculation ratios typically vary from 0.5 to 4.

Recirculation is beneﬁcial for the following reasons:

•

It improves the removal efﬁciency by increasing the contact time of the zoogleal layer with the

wastewater

• During low ﬂows it prevents the trickling ﬁlter from drying out

• It dilutes any toxic loadings

• It promotes oxygen transfer and reduces ponding

•

The increased hydraulic loading promotes uniform sloughing, prevents ponding, improves ventila-

tion through the ﬁlter and reduces potential for snail and ﬁlter ﬂy breeding

9.3 Operation of Multiple Trickling Filters

127

9.3 Operation of Multiple Trickling Filters

Multiple trickling ﬁlters can be operated in series or in parallel:

• Series operation in which the ﬂow from one ﬂows into the next.

– For high strength loading and for nitriﬁcation

• Parallel operation in which the trickling ﬁlters that are operated side by side.

– For winter operation - prevents freezing in the TF

9.4 Parameters for monitoring and operating trickling ﬁlters

1. Hydraulic loading is expressed as gpd/ ft²

2. BOD Removal (%)

3. Organic loading lbs BOD/day/ 1000 cu ft

4. Recirculation ratio

9.5 Classifying Trickling Filters

Trickling ﬁlters are classiﬁed according to the hydraulic and organic loading applied to the ﬁlter

9.5.1 Low-rate ﬁlter

•

The standard rate or low rate trickling ﬁlters (LRTF) are relatively simple treatment units that

normally produce a consistent efﬂuent quality even with varying inﬂuent strength

• They are generally not provided with recirculation of efﬂuent

•

Depending upon the dosing system, wastewater is applied intermittently with rest periods which

generally do not exceed ﬁve minutes at the designed rate of waste ﬂow.

While there is some unloading or sloughing of solids at all times, the major unloadings usually

occur several times a year for comparatively short Periods of time.

•

Hydraulic loading is 25 - 100 gal/day/sq. ft

BOD Removal (%) 50 – 80%

Organic loading is 5 - 25 lbs BOD/day/1000 cu ft

9.5.2 High-rate ﬁlter

The most important element of a high-rate trickling ﬁlter is the provision where a part of the settled

treated efﬂuent is pumped to the PST or to the ﬁlter. This is termed as Recirculation

•

128

Chapter 9. Trickling Filters

•

High-rate ﬁlters are usually characterized by higher hydraulic and organic loadings than low-rate

ﬁlters

The higher BOD loading is accomplished by applying a larger volume of waste per unit surface

area of the ﬁlter.

As a result of the higher ﬂow velocities a more continuous and uniform sloughing of excess

zoogleal growth occurs

Hydraulic loading is 100 - 1000 gal/day/sq. ft

BOD Removal (%) 65 - 85%

Organic loading is 25 - 100 lbs BOD/day/ 1000 cu ft

9.5.3 Roughing ﬁlter

• Roughing ﬁlters are high rate type ﬁlters designed with plastic packing

• In most cases roughing ﬁlters are used to treat wastewater prior to secondary treatment

•

One of the advantages of roughing ﬁlter is low energy requirement for BOD removal of high

strength wastewaters as compared to activated sludge process because the energy required is only

for pumping the inﬂuent wastewater and recirculation ﬂows

9.6 Trickling Filter Operational Issues

9.6.1 Ponding

If the voids in the media get plugged, ﬂow can collect on the surface in ponds. Correction: spraying the

surface with high pressure water stream stopping a rotary distributor over the ponded area hand-stir the

media or open the voids dose the ﬁlter with chlorine for several hours

9.6.2 Odors

• Corrective measures should be taken immediately if foul odors develop

• presence of foul odors indicates anaerobic conditions are predominant

• Check the under drain system for obstructions or heavy biological growths

• increase the recirculation rate to provide more oxygen to the ﬁlter bed and increase sloughing

• keep slime growths off of sidewalks and inside walls of the ﬁlter to reduce the odor

9.6.3 Trickling Filter Flies - Psychoda

• tiny, gnat-size ﬁlter ﬂy, or Psychoda - primary nuisance insect

• Correction methods include

– Increase recirculation rate

– keep oriﬁce openings clear

– apply insecticides to ﬁlter walls

– dose ﬁlter with chlorine

– keep weeds and tall grass cut around ﬁlter

9.6.4 Cold weather problems

• ice can form on the media of the ﬁlter

• Correction methods include:

– decrease recirculation to the ﬁlter (inﬂuent is usually warmer than recycled ﬂows)

– construct wind screens

9.6 Trickling Filter Operational Issues

129

– operate two-stage ﬁlters in parallel rather than in series

Chapter Assessment

1. Compared to activated sludge, trickling ﬁlters are sturdy work units not easily up set by shock

loads.

*a. True

b. False

2. Ponding on the surface of trickling ﬁlters is almost always caused by plugging of the underdrains

immediately below the point of ponding

a. True

*b. False

3. Zoogleal mass sheared of from the trickling ﬁlter media and carried out as part of the efﬂuent ﬂow

is termed as sloughing

*a. True

b. False

4. For multiple trickling ﬁlter operation, the parallel mode is more suitable than the series mode for

treating inﬂuent wastewater with a high BOD content

a. True

*b. False

5. Synthetic /pre-manufactured media used in the bioﬁlter has the advantage of providing a greater

surface area per volume upon which the zoologeal ﬁlm may grow while providing ample void

space for the free circulation of air.

*a. True

b. False

6. Of the two forms of secondary treatment: trickling ﬁlters and conventional activated sludge, trick-

ling ﬁlters are more likely to produce a clearer efﬂuent.

a. True

*b. False

7. A trickling ﬁlter reduces the strength of wastewater applied to it by the ﬁltering and straining action

132

Chapter 9. Trickling Filters

of the stones or support media.

a. True

*b. False

8. A trickling ﬁlter provides higher BOD removal in cold weather than in hot weather.

a. True

*b. False

9. The biological growth on trickling ﬁlter media is composed principally of aerobic forms of bacteria,

fungi, algae, protozoa, worms and the larvae of insects.

a. True

*b. False

10. The spaces between the ﬁlter media in a trickling ﬁlter need to remain open to permit free ﬂow of

air throughout the bed.

*a. True

b. False

11. The ﬂooding of ﬁlter media with wastewater is sometimes effective in killing ﬁlter ﬂies.

*a. True

b. False

12. Sandstone, because of its ability to provide excellent ventilation usually is used for media in trick-

ling ﬁlters.

a. True

*b. False

13. When a trickling ﬁlter is referred to as a "roughing ﬁlter", it means the operational life of the ﬁlter

is at an end, and a major overhaul is eminent.

a. True

*b. False

14. The hydraulic loading applied. to a trickling ﬁlter is the total volume of liquid, including recircula-

tion.

*a. True

b. False

15. The main function of a launder in a secondary clariﬁer is to prevent scum and other ﬂoatables from

leaving with the efﬂuent ﬂow

a. True

*b. False

16. The trickling ﬁlter recirculation ratio is calculated as Qi/Qr where Qi is the inﬂuent ﬂow and Qr is

the recirculated ﬂow.

a. True

*b. False

17. The advantage of trickling ﬁlter over the activated sludge lies in its BOD removal efﬁciency

a. True

*b. False

18. Recirculation helps prevent excessive sloughing

a. True

*b. False

19. Series operation of trickling ﬁlter is the preferred mode during very low temperature conditions

during winter

9.6 Trickling Filter Operational Issues

133

a. True

*b. False

20. A good reason to run a trickling ﬁlter in the series mode of operation is:

a. During cold weather to prevent ice formation

b. When the inﬂuent BOD loading is low

*c. When the inﬂuent BOD loading is high

d. When the weather is warm

21. A good reason to run a trickling ﬁlter in the parallel mode of operation is:

a. When the weather is warm

b. When the BOD loading is high

*c. When the BOD loading is low

d. To prevent ﬁlter ﬂies

22. A pan test should be done monthly on a trickling ﬁlter to:

a. Check oil quality in the distributor bearing

*b. Check sewage distribution on ﬁlter

c. To check ﬂow rates through the media

d. All the above

e. None of the above

23. A problem associated with trickling ﬁlters is

a. Bulking

b. Protozoans

*c. Ponding

d. Liquefaction

24. A trickling ﬁlter or activated sludge process causes nitriﬁcation. which is

a. Conversion of nitrogen to nitrate

b. Conversion of nitrogen to ammonia

c. Conversion of nitrate to nitrogen

*d. Conversion of ammonia to nitrate and nitrite nitrogen

25. If a trickling ﬁlter has been operating with a hydraulic loading (including some recirculation) of

10 to 12 MGD and organic loading of about 80 pounds of BOD per 1,000 cu. ft./day the treatment

efﬁciency will usually increase if the recirculation is increased. This might be attributed to:

a. The increased recirculation wears down the soluble BOD to ﬁner particles

*b. The increased ﬂow more completely wets and contacts all of the slime surfaces in the ﬁlter so

the food to effective microorganism ration is less as in an activated sludge process

c. The grazing population of warm and other organisms in the ﬁlter is ﬂushed out before they

consume the slime bacteria

d. The increased ﬂow more completely ﬁlls the under drains so cold updrafts are eliminated

e. The statement is just poppycock put out by power companies to get us to use more electricity to

run pumps

26. Most common trickling ﬁlter operational control method is:

a. Sloughing control

*b. Recirculation

c. Sludge removal

d. Distributor arm speed

27. The media in trickling ﬁlters is placed

134

Chapter 9. Trickling Filters

a. On a rubber tile ﬂoor.

*b. On a system of tile or columnar underdrains.

c. Directly in a concrete slab.

d. Directly in the ground

28. Sloughing from a trickling ﬁlter refers to

a. A process whereby wastewater is recirculated over the ﬁlter.

b. Bypassing of the ﬁlter.

c. The breaking of the ﬁlter stone as a result of weathering and the sluicing of small stone particles

to the ﬁnal settling tank.

*d. The periodic discharge of large quantities of slime growth with the ﬁlter efﬂuent.

29. The primary organisms responsible for treating wastewater in the trickling ﬁlter process are:

a. Anaerobic bacteria

b. Anoxic bacteria

c. Facultative bacteria

*d. Aerobic bacteria

30. The primary organisms responsible for treating wastewater in the trickling ﬁlter process are:

a. Anaerobic bacteria

b. Anoxic bacteria

c. Facultative bacteria

*d. Aerobic bacteria

31. Ponding of a trickling ﬁlter means

a. Flooding the ﬁlter.

b. Hosing the growth from the rocks with a high pressure hose.

*c. Blockage in the media that s prevents water to ﬂow through.

d. Running the trickling ﬁlter efﬂuent to a wastewater pond.

32. Which of the following would be a cause for trickling ﬁlter ponding:

a. Increased recirculation

b. Filter ﬂies

c. Zoogleal mass

*d. An excessive organic loading without a corresponding high recirculation rate

33. A good reason to run a trickling ﬁlter in the series mode of operation is:

a. During cold weather to prevent ice formation

b. When the inﬂuent BOD loading is low

*c. When the inﬂuent BOD loading is high

d. When the weather is warm

34. A good reason to run a trickling ﬁlter in the parallel mode of operation is:

a. When the weather is warm

b. When the BOD loading is high

*c. When the BOD loading is low

d. To prevent ﬁlter ﬂies

35. A problem associated with trickling ﬁlters is

a. Bulking

b. Protozoans

*c. Ponding

d. Liquefaction

9.6 Trickling Filter Operational Issues

135

36. A trickling ﬁlter or activated sludge process causes nitriﬁcation. which is

a. Conversion of nitrogen to nitrate

b. Conversion of nitrogen to ammonia

c. Conversion of nitrate to nitrogen

*d. Conversion of ammonia to nitrate and nitrite nitrogen

37. Which test best measures the efﬁciency of a trickling ﬁlter?

a. Total solids.

b. pH

*c. BOD

d. Temperature

e. Sludge age

38. What are sloughings?

a. Troughs.

b. Slop.

*c. Zoogleal mass washed off trickling ﬁlter media.

d. Waste-activated sludge.

e. Grit.

39. Most common trickling ﬁlter operational control method is:

a. Sloughing control

*b. Recirculation

c. Sludge removal

d. Distributor arm speed

40. Most common trickling ﬁlter operational control method is:

a. Sloughing control

*b. Recirculation

c. Sludge removal

d. Distributor arm speed

41. The rotating biological contactors operate based upon the same principles as:

a. Aerated lagoons

b. Activated sludge systems

*c. Trickling ﬁlters

d. Extended aeration

42. A two-stage trickling ﬁltration system means:

a. one ﬁlter used with recirculation returning from ﬁlter efﬂuent to ﬁlter inﬂuent at 100b. one ﬁlter

used with recirculation from secondary clariﬁer efﬂuent to trickling ﬁlter inﬂuent

*c. two ﬁlters used in series, either directly or with a clariﬁer in between

d. two ﬁlters used in parallel

e. one ﬁlter used with no primary clariﬁcation

43. Regardless of shape, one acre-foot of media in a trickling ﬁlter is equal to:

a. 33,000 cu. ft.

*b. 43,560 cu. ft.

c. 55,560 cu. ft.

d. 77,840 cu. ft.

e. none of the above

44. The presence of a "rotten egg" odor in the area of a trickling ﬁlter generally would indicate which

136

Chapter 9. Trickling Filters

of the following:

a. the presence of the psychoda ﬂy

b. the clogging of the distributor arm oriﬁces

*c. anaerobic conditions within the ﬁlter

d. too much recirculation

e. too high a DO in the: wastewater being applied to the ﬁlter

45. Analysis of a wastewater efﬂuent from a standard rate trickling ﬁlter shows the following:

Nitrite= 8 mg/L

Ammonia= 2 mg/L

Nitrate= 2 mg/L

What do these results indicate?

*a. Plant is operating properly

b. Final end product, nitrite, is too low

c. Final end product, nitrate, is too low

d. DO was insufﬁcient to allow complete aerobic oxidation of the nitrate to nitrite

46. Given the data below, what is the most likely cause of the trickling ﬁlter problem? –

DATA: Normal dry weather plant ﬂow rate

Sweep arms operating at normal rate

Water not passing through as rapidly as normal

Increased odor level

Increased chlorine demand in ﬁnal efﬂuenTrue

Final efﬂuent turbid.

*a. Filter media plugged with dead microorganisms

b. Low solids load to ﬁlter

c. seal in sweep arms worn

d. Spray nozzles partially plugged.

47. The rock/media in most trickling ﬁlters is placed

a. Directly in the ground.

b. Directly in a concrete slab.

*c. On a system of tile underdrains

d. On a rubber tile ﬂoor

48. The primary organisms responsible for treating wastewater in the trickling ﬁlter process are:

a. Anaerobic bacteria

b. Anoxic bacteria

c. Facultative bacteria

*d. Aerobic bacteria

49. Trickling ﬁlters usually consist of a bed of stone which performs its function in sewage by:

a. Mechanical ﬁltration of organic solids

b. Mechanical ﬁltration of inorganic solids

c. Provides aeration

*d. Supports a growth of organisms which feed upon the sewage and reduce BOD

e. None of the above

50. What are sloughings?

a. Troughs.

b. Slop.

9.6 Trickling Filter Operational Issues

137

*c. Zoogleal mass washed off trickling ﬁlter media.

d. Waste-activated sludge.

e. Grit.

51. Which of the following is not a beneﬁt of recirculation in the trickling ﬁlter process?

a. Dilutes high strength or toxic wastes.

b. Helps prevent septic conditions in trickling ﬁlter

*c. Helps prevent excessive sloughing

d. Helps control odors, ponding and ﬁlter ﬂies

10. Stabilzation Ponds

•

Stabilization ponds and lagoons are bodies of water which treat wastewater using mainly natural

processes including sunlight, algae and microorganisms for treating wastewater

• While ponds are shallow and man-made, lagoons are bodies of water conﬁned within natural bound-

aries.

10.1 Advantages of ponds

• Cheap to build and operate

• Low maintenance and electrical costs

• Do not require highly trained operational personnel

•

Provide treatment that can be equal to some secondary treatment processes and have fewer sludge

handling issues.

10.2 Disadvantages of ponds

• Land intensive

• Efﬂuent quality varies with seasonal temperature changes

• Suspended solids levels that can create regulatory problems.

• System upsets almost always result in odor problems and recovery times may be weeks or months.

• Not appropriate for colder climates

10.3 Types of Ponds

10.3.1 Anaerobic Ponds

Anaerobic Ponds

• Typically for treating raw sewage

140

Chapter 10. Stabilzation Ponds

•

These are deep - 10-14 feet treatment ponds which rely primarily on anaerobic bacteria to break

down the organic waste.

• Designed for BOD removal.

• High strength wastewater may be treated.

•

Organic matter is broken down releasing releasing methane, carbon dioxide and odorous gases

including hydrogen sulﬁde.

• Most of the decomposition is accomplished by acid forming bacteria.

• The pH in these lagoons is usually below 6.5.

• They are total retention and do not have an efﬂuent discharge.

• The anaerobic pond must be de-sludged approximately once every 2 to 5 years

• Organic loading of 200-1000 lbs. BOD₅per acre per day

10.3.2 Facultative Ponds

The depth of facultative ponds is about 4-7 feet which is in-between the depths of anaerobic ponds

(10-14 feet) and aerobic ponds 3 feet)

• The uper layer of facultative pond is aerobic, and bottom layer is mostly anaerobic.

•

Facultative bacteria are responsible for most of the treatment that occurs in these ponds. Facultative

bacteria are bacteria which can live under both aerobic and anaerobic conditions.

• The algae that grow in the pond are critical to the successful stabilization of the organic load.

•

The algae will take in carbon dioxide (CO₂) and, through photosynthesis, use it to create sugars and

release dissolved oxygen (O₂) that is used by the aerobic bacteria. Facultative lagoon levels should

always maintain at least 4 feet of water in the pond.

• Typically for secondary treatment - BOD removal

• 15-50 lbs BOD₅per acre per day.

•

Unused CO₂will react with water to form carbonic acid - which would reduce the pH unless con-

sumed

Sludge removal need is rare. Sludge can be removed by using a raft-mounted sludge pump or by

draining and dewatering the pond and removing the sludge with a front-end loader.

10.3.3 Aerobic Stabilization Ponds

Aerobic stabilization ponds are also known as: maturation, polishing or ﬁnishing Pond

• Contain disssolved oxygen throughout entire depth of the pond.

•

Treatment is accomplished through the stabilization of organic wastes by aerobic bacteria and

algae.

• Typically for tertiary treatment

• Designed for pathogen removal

• Shallow - only about 3 feet deep.

• They are most often the ﬁnal cells in a multi-staged pond system

• They are also used as polishing ponds for tertiary treatment of trickling ﬁlter plant efﬂuent.

• Usually the efﬂuent is directed into a second pond where the sludge can settle

•

Their shallow depth allows sunlight to penetrate to the bottom of the pond to encourage algae

growth and aerobic conditions throughout the pond

•

The low solids loading found in these tertiary treatment applications means that these ponds nor-

mally have no sludge zone

• These ponds may be mechanically aerated

10.3 Types of Ponds

141

142

Chapter 10. Stabilzation Ponds

• Aerobic polishing ponds are designed for 15-20 pounds BOD/acre/day

• Aerobic ponds are typically designed for pathogen removal

• Aerobic lagoon levels should always maintain at least 18 inches of water in the pond

10.4 Ponds Operations and Maintenance

• Ponds and lagoons are designed as continuous discharge, controlled discharge, or no discharge.

–

In controlled discharge ponds the wastewater is held for long periods of time before discharg-

ing.

–

No-discharge ponds the inﬂow rate needs to be equal or exceed the rates of evaporation

and/or percolation.

•

Short-circuiting in a pond may be caused by poor design of inlet and outlet piping arrangements or

by uncontrolled growth of water weeds.

Stagnant water will breed mosquitoes and can result in anaerobic conditions developing that can

cause odor issues

• Dikes need to be maintained

• Aquatic plants and weeds must be removed from the water.

• Reeds will create stagnant areas along the edge of the pond and need to be removed

• To start a new pond, two feet of water is typically added prior to fresh starting wastewater feed.

•

Sodium nitrate can be used to help recover from an odor-causing upset. The nitrates (NO₃) will

provide a source of chemically bound oxygen for the bacteria to use instead of dissolved oxygen.

• Scum control may be required

Chapter Assessment

1. Organic loading to a stabilization pond is always calculated by knowing the pounds of BOD ap-

plied per 1.000 cubic feet of pond volume per day.

a. True

*b. False

2. An abundance of aquatic plant growth in and around the edge of a facultative pond provides more

surface area for biologic activity and increases treatment capacity.

a. True

*b. False

3. Facultative ponds are anaerobic on the bottom and aerobic near the surface.

*a. True

b. False

4. The best time of the year to initiate operation of a new facultative pond is during the coldest months

of the year.

a. True

*b. False

5. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

6. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

7. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

144

Chapter 10. Stabilzation Ponds

8. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

9. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

10. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

11. Facultative ponds are anaerobic on the bottom and aerobic near the surface.

*a. True

b. False

12. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

b. False

13. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

14. Algae produce oxygen in a facultative pond.

*a. True

b.False

15. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

16. In a facultative pond, aerobic bacteria produce oxygen that is consumed by algae.

a. True

*b. False

17. In a facultative pond, the pH falls when carbon dioxide produced by the aerobic bacteria is con-

sumed by the algae

a. True

*b. False

18. In a facultative pond, the algae utilizes the oxygen given off as a by-product of bacterial respiration

a. True

*b. False

19. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

20. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

b. False

10.4 Ponds Operations and Maintenance

145

21. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

22. In a facultative wastewater treatment pond that Is operating normally, a rise In DO will be accompa-

nied by a drop in pH.

a. True

*b. False

23. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

24. Operational depth of a wastewater pond is not used for calculating its organic loading

*a. True

b. False

25. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

26. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

27. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

28. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

29. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

30. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

31. The best time of the year to initiate operation of a new facultative pond is during the coldest months

of the year.

a. True

*b. False

32. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

33. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

146

Chapter 10. Stabilzation Ponds

a. True

*b. False

34. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

35. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

36. A wastewater pond may have a detention time of 30 to 120 days.

*a. True

b. False

37. Sludge deposits are usually in evidence at the efﬂuent end of a waste treatment pond rather than the

inﬂuent end.

a. True

*b. False

38. It is good practice to start waste treatment ponds during winter because of the lower level of biolog-

ical activity.

a. True

*b. False

39. Anaerobic ponds require higher. minimum temperatures than aerobic ponds to achieve satisfactory

treatment,

*a. True

b. False

40. The formula for calculating percent BOD removal in pond systems is: BOD removal,

*a. True

b. False

41. The organic loading on a wastewater lagoon is higher than on a polishing pond.

*a. True

b. False

42. Algae produce oxygen in a facultative pond.

*a. True

b. False

43. A disadvantage of excessive scum on an oxidation pond is that it will interfere with photosynthesis.

*a. True

b. False

44. Ranges of treatment efﬁciencies that can be expected from ponds vary more than most other treat-

ment processes.

*a. True

b. False

45. At night, algae in a facultative pond consume oxygen.

*a. True

b. False

46. In wastewater treatment ponds, DO levels very rarely exceed 15 mg/l at the pond surface.

10.4 Ponds Operations and Maintenance

147

a. True

*b. False

47. Organic loading to a wastewater treatment pond is expressed in units of pounds of VSS per acre per

day.

a. True

*b. False

48. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

49. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

50. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

51. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

b. False

52. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

53. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

54. Algae produce oxygen in a facultative pond.

*a. True

b.False

55. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

56. In a facultative pond, aerobic bacteria produce oxygen that is consumed by algae.

a. True

*b. False

57. In a facultative pond, the pH falls when carbon dioxide produced by the aerobic bacteria is con-

sumed by the algae

a. True

*b. False

58. In a facultative pond, the algae utilizes the oxygen given off as a by-product of bacterial respiration

a. True

*b. False

59. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

148

Chapter 10. Stabilzation Ponds

*b. False

60. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

b. False

61. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

62. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

a. True

*b. False

63. In a facultative wastewater treatment pond that Is operating normally, a rise In DO will be accompa-

nied by a drop in pH.

a. True

*b. False

64. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

65. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

66. Operational depth of a wastewater pond is not used for calculating its organic loading

*a. True

b. False

67. The best time of the year to initiate operation of a new facultative pond is during the coldest months

of the year.

a. True

*b. False

68. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

69. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

70. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

71. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

10.4 Ponds Operations and Maintenance

149

*a. True

b. False

72. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

73. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

74. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

75. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

76. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

b. False

77. Ranges of treatment efﬁciencies that can be expected from ponds vary more than most other treat-

ment processes.

*a. True

b. False

78. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

79. The best time of the year to initiate operation of a new facultative pond is during the coldest months

of the year.

a. True

*b. False

80. The color of the algae at the surface of a facultative pond can be an indication of its operational

condition.

*a. True

b. False

81. The ﬂow to a wastewater treatment pond may be expressed in units of acre -feet.

a. True

*b. False

82. Usually facultative ponds are mixed with mechanical aerators at night and on cloudy days to dis-

perse food and bacteria evenly.

a. True

*b. False

83. Wastewater treatment ponds that receive untreated raw wastewater are called stabilization ponds.

*a. True

b. False

84. When a facultative pond develops a green appearance, it should be treated with copper sulfate to

kill off the algae.

150

Chapter 10. Stabilzation Ponds

a. True

*b. False

85. When operating ponds in series, the accumulation of solids in the ﬁrst pond may become a problem

after long periods of use.

*a. True

b. False

86. Organic loading to a stabilization pond is always calculated by knowing the pounds of BOD ap-

plied per 1.000 cubic feet of pond volume per day.

a. True

*b. False

87. An abundance of aquatic plant growth in and around the edge of a facultative pond provides more

surface area for biologic activity and increases treatment capacity.

a. True

*b. False

88. Facultative ponds are anaerobic on the bottom and aerobic near the surface.

*a. True

b. False

89. The best time of the year to initiate operation of a new facultative pond is during the coldest months

of the year.

a. True

*b. False

90. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.

*a. True

b. False

91. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen

than the surface layer.

a. True

*b. False

92. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-

lems and to take advantage of the increased solubility of oxygen in cold water.

a. True

*b. False

93. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million

gallons.

*a. True

b. False

94. Algae is primarily responsible for the effective working of an anaerobic pond

a. True

*b. False

95. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream

a. True

*b. False

96. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved

oxygen.

*a. True

10.4 Ponds Operations and Maintenance

151

b. False

97. One short-term corrective measure for an overloaded facultative pond might be to add:

a. copper sulfate .

b. sodium sulﬁde.

c. ammonium sulﬁde .

*d. sodium nitrate.

e. potassium chloride.

98. Photosynthesis is an essential part of the biological activity associated with:

a. Activated sludge

b. Trickling ﬁlters

c. Oxidation ditches

d. Aerobic digesters

*e. Sewage lagoons

99. During the process of algal photosynthesis:

a. Chlorophyll converts sunlight into energy for growth.

b. Algae produces oxygen

c. Algae converts CO2, NH3, and PO4, into additional algae cells

*d. All of the above

100. pH of the facultative pond will be the highest

*a. during daytime when the consumption of CO2 is the highest

b. during daytime when the consumption of CO2 is the lowest

c. during nighttime when the production of CO2 is highest

d. during nighttime when the production of CO2 is lowest

101. A lagoon operator collects a sample of efﬂuent at 2:15 pm. on a sunny July day and tests it. for

dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green

color. Efﬂuent suspended solids have been running at 75 mg/1. The operator should

a. Do nothing. The conditions described are normal.

*b. Apply algaecide to the lagoon to kill the algae.

c. Drawdown the lagoon to eliminate excess DO.

d. Isolate cell.

102. Algae in a stabilization pond is most likely to:

a. consume oxygen during daylight hours.

b. decrease efﬂuent TSS during the day.

*c. change the pH throughout the day.

d. increase oxygen at night.

e. None of the above.

103. Algae in a stabilization pond is most likely to:

a. consume oxygen during daylight hours.

b. decrease efﬂuent TSS during the day.

*c. change the pH throughout the day.

d. increase oxygen at night.

e. None of the above.

104. An operator cannot maintain adequate water levels in one of the ponds. What will cause this to

happen?

a. The pond is hydraulically overloaded.

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Chapter 10. Stabilzation Ponds

*b. The pond seal leaks.

c. The ﬂow control structure leaks.

d. The ﬂow control structure does not split the ﬂow evenly.

105. A properly designed and operated wastewater stabilization pond will remove

percent

BOD.

a. 40%-60%.

b. 60%-70%.

c. 70%-80%.

*d. 80%-90%.

106. At night algae in a conventional lagoon will:

*a. Cease to produce oxygen

b. Consume oxygen

c. Produce less oxygen

d. Increase the pH of the lagoon contents

107. At what time of day is the dissolved oxygen content highest in a lagoon?

a. 3 a.m.

b. 7 a.m.

c. 9 a.m.

*d. 3 p.m.

108. Cattails growing in lagoon will

*a. Cause short circuiting in affected lagoon.

b. Eliminate mosquito larvae.

c. Increase the diurnal pH ﬂuctuations.

d. Increase toxic blue-green algae concentrations in the efﬂuent.

109. Dike vegetation should be controlled by

*a. Mowing periodically.

b. Burning in the spring and fall.

c. Allowing the cattle to graze on the dikes.

d. Any of the above would be acceptable.

110. Due to diurnal differences in operation, a lagoon system is likely to experience the lowest dissolved

oxygen readings

a. At any time since diurnal differences have no bearing on DO values

b. During the time when the sun is out and it is the hottest

*c. During night, just before dawn

d. When BOD loading is the lowest

111. During the process of algal photosynthesis:

a. Chlorophyll converts sunlight into energy for growth.

b. Algae produces oxygen

c. Algae converts CO2, NH3, and PO4, into additional algae cells

*d. All of the above

112. Given the following data, what is the most likely cause of the aerated pond problem? — DATA:

Turbulence over center part of pond. Solids ﬂoating to surface on edge of pond. DO in center

portion= 1.4 mg/L.

a. Hydraulic ﬂow too slow through pond.

*b. Inadequate aeration on edges on pond.

10.4 Ponds Operations and Maintenance

153

c. Sludge scraper not operating.

d. Toxic materials in pond.

113. Hydraulic loading to a facultative pond equals:

a. Volume (MG) divided by Flow (MGD)

*b. Depth (inches) divided by Detention Time (days)

c. Volume (acre-feet) divided by Flow (acre-inches/day)

d. Flow ( gallons/day) divided by Area (acres)

e. Area (acres) divided by Flow (acre-inches/day)

114. In a properly operating facultative pond, algae live on carbon dioxide and nutrients during the day,

and at night produce carbon dioxide. This has what effect on the pH?

*a. pH increases during the day, and decreases at night

b. pH decreases during the day, and increases at night

c. pH stays the same no matter what time of day

d. Carbon dioxide has no effect on pH

115. It is important to completely mix and aerate the following type of pond:

*a. aerobic.

b. facultative.

c. anaerobic.

d. All of the above.

e. None of the above.

116. One short-term corrective measure for an overloaded facultative pond might be to add:

a. copper sulfate .

b. sodium sulﬁde.

c. ammonium sulﬁde .

*d. sodium nitrate.

e. potassium chloride.

117. pH of the facultative pond will be the highest

*a. during daytime when the consumption of CO2 is the highest

b. during daytime when the consumption of CO2 is the lowest

c. during nighttime when the production of CO2 is highest

d. during nighttime when the production of CO2 is lowest

118. Photosynthesis is an essential part of the biological activity associated with:

a. Activated sludge

b. Trickling ﬁlters

c. Oxidation ditches

d. Aerobic digesters

*e. Sewage lagoons

119. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-

solved oxygen readings

a. At any time since diurnal differences have no bearing on DO values

b. During the time when the sun is out and it is the hottest

*c. During night, just before dawn

d. When BOD loading is the lowest

120. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-

solved oxygen readings

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Chapter 10. Stabilzation Ponds

a. At any time since diurnal differences have no bearing on DO values

b. During the time when the sun is out and it is the hottest

*c. During night, just before dawn

d. When BOD loading is the lowest

121. The main function of algae in a facultative wastewater treatment pond is to:

a. produce carbon dioxide, which is then used by the facultative bacteria.

b. use up nutrients such as nitrogen and phosphorus.

*c. produce oxygen during daylight hours.

d. serve as a food source for essential protozoa

e. break down complex organics in the wastewater.

122. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-

tive wastewater treatment pond?

a. Aerobic bacteria

b. Anaerobic

c. Facultative bacteria

*d. All of the above

e. (a) and (c) only

123. Which of the following terms is usually not associated with wastewater treatment ponds?

*a. Filamentous bacteria

b. Symbiotic relationship

c. Photosynthesis

d. Sewage lagoon

e. Inches/day

124. Which of the following statements is not true regarding a facultative wastewater treatment pond?

a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond

prior to the addition of wastewater.

b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.

*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during

the afternoon of a sunny day.

d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.

e. A facultative pond has an anaerobic layer and an aerobic layer.

125. The main function of algae in a facultative wastewater treatment pond is to:

a. produce carbon dioxide, which is then used by the facultative bacteria.

b. use up nutrients such as nitrogen and phosphorus.

*c. produce oxygen during daylight hours.

d. serve as a food source for essential protozoa

e. break down complex organics in the wastewater.

126. Which of the following terms is usually not associated with wastewater treatment ponds?

*a. Filamentous bacteria

b. Symbiotic relationship

c. Photosynthesis

d. Sewage lagoon

e. Inches/day

127. Which of the following statements is not true regarding a facultative wastewater treatment pond?

a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond

10.4 Ponds Operations and Maintenance

155

prior to the addition of wastewater.

b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.

*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during

the afternoon of a sunny day.

d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.

e. A facultative pond has an anaerobic layer and an aerobic layer.

128. Which of the following would not be a routine operational or maintenance problem in the operation

of pond:

a. weed control.

b. levee maintenance.

c. insect control.

*d. temperature control.

e. scum control.

129. Algae in a stabilization pond is most likely to:

a. consume oxygen during daylight hours.

b. decrease efﬂuent TSS during the day.

*c. change the pH throughout the day.

d. increase oxygen at night.

e. None of the above.

130. At night algae in a conventional lagoon will:

*a. Cease to produce oxygen

b. Consume oxygen

c. Produce less oxygen

d. Increase the pH of the lagoon contents

131. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-

tive wastewater treatment pond?

a. Aerobic bacteria

b. Anaerobic

c. Facultative bacteria

*d. All of the above

e. (a) and (c) only

132. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?

a. Oxidation pond

b. Stabilization pond

c. Facultative lagoon

*d. Aerobic lagoon

133. The main function of algae in a facultative wastewater treatment pond is to:

a. produce carbon dioxide, which is then used by the facultative bacteria.

b. use up nutrients such as nitrogen and phosphorus.

*c. produce oxygen during daylight hours.

d. serve as a food source for essential protozoa.

e. break down complex organics in the wastewater.

134. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore

normal conditions after a pond "turn over." Its function is:

*a. To provide a source of chemically combined oxygen for the facultative bacteria.

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Chapter 10. Stabilzation Ponds

b. To kill off the unwanted blue-green algae.

c. To preserve the nitrogen balance in the pond.

d. To control the growth of weeds.

e. To neutralize chlorine residuals in the pond efﬂuent.

135. It is important to completely mix and aerate the following type of pond:

*a. aerobic.

b. facultative.

c. anaerobic.

d. All of the above.

e. None of the above.

136. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:

a. pH, color, and BOD

*b. pH, BOD, and suspended solids

c. pH, DO, and temperature

d. DO, BOD, and SS

e. Microscopic examination, color, and DO

137. The efﬂuent from a conventional lagoon should be withdrawn:

a. Off the surface

b. Near the bottom

*c. Six to eighteen inches below the surface

d. Intermittently

138. Algae in a stabilization pond is most likely to:

a. consume oxygen during daylight hours.

b. decrease efﬂuent TSS during the day.

*c. change the pH throughout the day.

d. increase oxygen at night.

e. None of the above.

139. Which of the following terms is usually not associated with wastewater treatment ponds?

*a. Filamentous bacteria

b. Symbiotic relationship

c. Photosynthesis

d. Sewage lagoon

e. Inches/day

140. Which of the following statements is not true regarding a facultative wastewater treatment pond?

a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond

prior to the addition of wastewater.

b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.

*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during

the afternoon of a sunny day.

d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.

e. A facultative pond has an anaerobic layer and an aerobic layer.

141. One short-term corrective measure for an overloaded facultative pond might be to add:

a. copper sulfate .

b. sodium sulﬁde.

c. ammonium sulﬁde .

10.4 Ponds Operations and Maintenance

157

*d. sodium nitrate.

e. potassium chloride.

142. Photosynthesis is an essential part of the biological activity associated with:

a. Activated sludge

b. Trickling ﬁlters

c. Oxidation ditches

d. Aerobic digesters

*e. Sewage lagoons

143. pH of the facultative pond will be the highest

*a. during daytime when the consumption of CO2 is the highest

b. during daytime when the consumption of CO2 is the lowest

c. during nighttime when the production of CO2 is highest

d. during nighttime when the production of CO2 is lowest

144. One short-term corrective measure for an overloaded facultative pond might be to add:

a. copper sulfate .

b. sodium sulﬁde.

c. ammonium sulﬁde .

*d. sodium nitrate.

e. potassium chloride.

145. Photosynthesis is an essential part of the biological activity associated with:

a. Activated sludge

b. Trickling ﬁlters

c. Oxidation ditches

d. Aerobic digesters

*e. Sewage lagoons

146. During the process of algal photosynthesis:

a. Chlorophyll converts sunlight into energy for growth.

b. Algae produces oxygen

c. Algae converts CO2, NH3, and PO4, into additional algae cells

*d. All of the above

147. pH of the facultative pond will be the highest

*a. during daytime when the consumption of CO2 is the highest

b. during daytime when the consumption of CO2 is the lowest

c. during nighttime when the production of CO2 is highest

d. during nighttime when the production of CO2 is lowest

148. A lagoon operator collects a sample of efﬂuent at 2:15 pm. on a sunny July day and tests it. for

dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green

color. Efﬂuent suspended solids have been running at 75 mg/1. The operator should

a. Do nothing. The conditions described are normal.

*b. Apply algaecide to the lagoon to kill the algae.

c. Drawdown the lagoon to eliminate excess DO.

d. Isolate cell.

149. Algae in a stabilization pond is most likely to:

a. consume oxygen during daylight hours.

b. decrease efﬂuent TSS during the day.

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Chapter 10. Stabilzation Ponds

*c. change the pH throughout the day.

d. increase oxygen at night.

e. None of the above.

150. An operator cannot maintain adequate water levels in one of the ponds. What will cause this to

happen?

a. The pond is hydraulically overloaded.

*b. The pond seal leaks.

c. The ﬂow control structure leaks.

d. The ﬂow control structure does not split the ﬂow evenly.

151. A properly designed and operated wastewater stabilization pond will remove what percent BOD.

a. 40%-60%.

b. 60%-70%.

c. 70%-80%.

*d. 80%-90%.

152. At night algae in a conventional lagoon will:

*a. Cease to produce oxygen

b. Consume oxygen

c. Produce less oxygen

d. Increase the pH of the lagoon contents

153. At what time of day is the dissolved oxygen content highest in a lagoon?

a. 3 a.m.

b. 7 a.m.

c. 9 a.m.

*d. 3 p.m.

154. Cattails growing in lagoon will

*a. Cause short circuiting in affected lagoon.

b. Eliminate mosquito larvae.

c. Increase the diurnal pH ﬂuctuations.

d. Increase toxic blue-green algae concentrations in the efﬂuent.

155. Dike vegetation should be controlled by

*a. Mowing periodically.

b. Burning in the spring and fall.

c. Allowing the cattle to graze on the dikes.

d. Any of the above would be acceptable.

156. Due to diurnal differences in operation, a lagoon system is likely to experience the lowest dissolved

oxygen readings

a. At any time since diurnal differences have no bearing on DO values

b. During the time when the sun is out and it is the hottest

*c. During night, just before dawn

d. When BOD loading is the lowest

157. Given the following data, what is the most likely cause of the aerated pond problem? —

DATA: Turbulence over center part of pond. Solids ﬂoating to surface on edge of pond. DO in

center portion= 1.4 mg/L.

a. Hydraulic ﬂow too slow through pond.

*b. Inadequate aeration on edges on pond.

10.4 Ponds Operations and Maintenance

159

c. Sludge scraper not operating.

d. Toxic materials in pond.

158. Hydraulic loading to a facultative pond equals:

a. Volume (MG) divided by Flow (MGD)

*b. Depth (inches) divided by Detention Time (days)

c. Volume (acre-feet) divided by Flow (acre-inches/day)

d. Flow ( gallons/day) divided by Area (acres)

e. Area (acres) divided by Flow (acre-inches/day)

159. In a properly operating facultative pond, algae live on carbon dioxide and nutrients during the day,

and at night produce carbon dioxide. This has what effect on the pH?

*a. pH increases during the day, and decreases at night

b. pH decreases during the day, and increases at night

c. pH stays the same no matter what time of day

d. Carbon dioxide has no effect on pH

160. It is important to completely mix and aerate the following type of pond:

*a. aerobic.

b. facultative.

c. anaerobic.

d. All of the above.

e. None of the above.

161. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-

solved oxygen readings

a. At any time since diurnal differences have no bearing on DO values

b. During the time when the sun is out and it is the hottest

*c. During night, just before dawn

d. When BOD loading is the lowest

162. The main function of algae in a facultative wastewater treatment pond is to:

a. produce carbon dioxide, which is then used by the facultative bacteria.

b. use up nutrients such as nitrogen and phosphorus.

*c. produce oxygen during daylight hours.

d. serve as a food source for essential protozoa

e. break down complex organics in the wastewater.

163. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-

tive wastewater treatment pond?

a. Aerobic bacteria

b. Anaerobic

c. Facultative bacteria

*d. All of the above

e. (a) and (c) only

164. Which of the following statements is not true regarding a facultative wastewater treatment pond?

a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond

prior to the addition of wastewater.

b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.

*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during

the afternoon of a sunny day.

160

Chapter 10. Stabilzation Ponds

d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.

e. A facultative pond has an anaerobic layer and an aerobic layer.

165. Which of the following would not be a routine operational or maintenance problem in the operation

of pond:

a. weed control.

b. levee maintenance.

c. insect control.

*d. temperature control.

e. scum control.

166. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?

a. Oxidation pond

b. Stabilization pond

c. Facultative lagoon

*d. Aerobic lagoon

167. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore

normal conditions after a pond "turn over." Its function is:

*a. To provide a source of chemically combined oxygen for the facultative bacteria.

b. To kill off the unwanted blue-green algae.

c. To preserve the nitrogen balance in the pond.

d. To control the growth of weeds.

e. To neutralize chlorine residuals in the pond efﬂuent.

168. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:

a. pH, color, and BOD

*b. pH, BOD, and suspended solids

c. pH, DO, and temperature

d. DO, BOD, and SS

e. Microscopic examination, color, and DO

169. The efﬂuent from a conventional lagoon should be withdrawn:

a. Off the surface

b. Near the bottom

*c. Six to eighteen inches below the surface

d. Intermittently

170. The facultative zone in a pond:

a. provides no treatment.

b. provides oxygen to the anaerobic zone.

c. is a barrier to light penetration

d. produces the bacteria for the anaerobic zone.

*e. uses nitrate as an oxygen source for digestion.

171. Which of the following terms is usually not associated with wastewater treatment ponds?

*a. Filamentous bacteria

b. Symbiotic relationship

c. Photosynthesis

d. Sewage lagoon

e. Inches/day

172. A lagoon operator collects a sample of efﬂuent at 2:15 pm. on a sunny July day and tests it. for

10.4 Ponds Operations and Maintenance

161

dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green

color. Efﬂuent suspended solids have been running at 75 mg/1. The operator should

a. Do nothing. The conditions described are normal.

*b. Apply algaecide to the lagoon to kill the algae.

c. Drawdown the lagoon to eliminate excess DO.

d. Isolate cell.

173. Given the following data, what is the most likely cause of the aerated pond problem? —

DATA: Turbulence over center part of pond. Solids ﬂoating to surface on edge of pond. DO in

center portion= 1.4 mg/L.

a. Hydraulic ﬂow too slow through pond.

*b. Inadequate aeration on edges on pond.

c. Sludge scraper not operating.

d. Toxic materials in pond.

174. Hydraulic loading to a facultative pond equals:

a. Volume (MG) divided by Flow (MGD)

*b. Depth (inches) divided by Detention Time (days)

c. Volume (acre-feet) divided by Flow (acre-inches/day)

d. Flow ( gallons/day) divided by Area (acres)

e. Area (acres) divided by Flow (acre-inches/day)

175. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?

a. Oxidation pond

b. Stabilization pond

c. Facultative lagoon

*d. Aerobic lagoon

176. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore

normal conditions after a pond "turn over." Its function is:

*a. To provide a source of chemically combined oxygen for the facultative bacteria.

b. To kill off the unwanted blue-green algae.

c. To preserve the nitrogen balance in the pond.

d. To control the growth of weeds.

e. To neutralize chlorine residuals in the pond efﬂuent.

177. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:

a. pH, color, and BOD

*b. pH, BOD, and suspended solids

c. pH, DO, and temperature

d. DO, BOD, and SS

e. Microscopic examination, color, and DO

178. The efﬂuent from a conventional lagoon should be withdrawn:

a. Off the surface

b. Near the bottom

*c. Six to eighteen inches below the surface

d. Intermittently

179. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore

normal conditions after a pond "turn over." Its function is:

*a. To provide a source of chemically combined oxygen for the facultative bacteria.

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Chapter 10. Stabilzation Ponds

b. To kill off the unwanted blue-green algae.

c. To preserve the nitrogen balance in the pond.

d. To control the growth of weeds.

e. To neutralize chlorine residuals in the pond efﬂuent.

180. The efﬂuent from a conventional lagoon should be withdrawn:

a. Off the surface

b. Near the bottom

*c. Six to eighteen inches below the surface

d. Intermittently

181. The facultative zone in a pond:

a. provides no treatment.

b. provides oxygen to the anaerobic zone.

c. is a barrier to light penetration.

d. produces the bacteria for the anaerobic zone.

*e. uses nitrate as an oxygen source for digestion.

182. One short-term corrective measure for an overloaded facultative pond might be to add:

a. copper sulfate .

b. sodium sulﬁde.

c. ammonium sulﬁde .

*d. sodium nitrate.

e. potassium chloride.

183. During the process of algal photosynthesis:

a. Chlorophyll converts sunlight into energy for growth.

b. Algae produces oxygen

c. Algae converts CO2, NH3, and PO4, into additional algae cells

*d. All of the above

184. pH of the facultative pond will be the highest

*a. during daytime when the consumption of CO2 is the highest

b. during daytime when the consumption of CO2 is the lowest

c. during nighttime when the production of CO2 is highest

d. during nighttime when the production of CO2 is lowest

11. Activated Sludge

11.1 Process overview

Activated sludge is a secondary, biological treatment process used for the removal of suspended and

dissolved organic matter from the primary treated wastewater.

• Utilizes an aeration basin/reactor and a secondary clariﬁer

•

In the presence of oxygen, aerobic bacteria in the aeration basin consume the organic matter (BOD)

in wastewater for their growth and reproduction, converting BOD into bacterial cell mass along

with metabolic byproducts including carbon dioxide and water

•

The aerobic bacteria is the predominant microbial life form in the aeration basin. Other higher

microbial life forms — mainly protozoa, are present along with some metazoans.

The microorganisms along with their metabolic byproducts and residual dead cell mass form a

cluster called ﬂoc.

The wastewater exiting the aeration basin enters a clariﬁer where the ﬂoc settles. The clear, treated

secondary efﬂuent ﬂows out.

A portion of the settled activated sludge ﬂoc, is returned from the clariﬁer to the front of the aera-

tion basin to seed the activated sludge treatment of the incoming primary efﬂuent. The recycled ﬂoc

is called Return Activated Sludge (RAS).

•

The remaining settled ﬂoc from the clariﬁer is ”wasted” —transferred for solids treatment (typ-

ically using digestion) prior to its ultimate disposal. The wasted ﬂoc is called Waste Activated

Sludge (WAS).

• The color of healthy activated sludge is tan to brown with an earthy/musty odor.

164

Chapter 11. Activated Sludge

•

For activated sludge treatment to be effective, it is critical to establish a healthy microbial popula-

tion which converts the BOD into easily separable biomass.

If the biomass does not settle well in the clariﬁer, it will be carried out in the treated secondary

efﬂuent producing a poor quality efﬂuent with higher solids and organic content.

Mixed liquor:

• Mixed liquor is the mixture of raw and/or treated wastewater with the biomass.

• Mixed liquor sample is typically taken at the end of the aeration reactor.

•

In the mixed liquor, the suspended solids comprised mostly of biomass with some undegraded and

non-degradable material is called Mixed liquor suspended solids (MLSS).

• The volatile fraction of the MLSS is termed Mixed Liquor Volatile Solids (MLVSS).

•

The MLVSS is the amount of biomass as a percentage of the MLSS. Generally, the MLVSS is 70

- 80%, implying that 70-80% of the MLSS is volatile (organic) solids.

11.2 Process parameters

11.2.1 Dissolved oxygen (DO)

Maintaining adequate Dissolved oxygen (DO) is key to the activated sludge process. DO is typically

maintained between 0.5 to 3.0 mg/l in the aeration reactor.

11.2.2 pH

The microbiological population in the activated sludge is very sensitive to pH and needs to be maintained

at neutral - 7 pH.

11.2.3 Temperature

Biological activity is temperature dependent. Microbial activity approximately doubles for every 10^oC

rise in temperature.

11.2.4 Oxygen uptake rate (OUR) and speciﬁc oxygen uptake rate (SOUR)

Oxygen Uptake Rate (OUR) involves measurement of the amount of oxygen used up by the microor-

ganisms using a DO probe and is expressed in unit time of mg/L-hr (ppm O₂consumed per hour). By

knowing the OUR, we can establish the activity of the microorganisms in the aeration tank and know if

they are consuming the oxygen provided for removing organic matter. For conventional activated sludge

process the typical OUR values range from 10 to 30 mg/L-hr

Speciﬁc Oxygen Uptake Rate (SOUR) provides the OUR information based on the concentration of

microorganisms present. SOUR is obtained by dividing OUR with MLVSS. The value is indicated and

11.2 Process parameters

165

O₂

l −hr

measured in terms of unit of

∗1000

MLVSS

Optimal range of SOUR is usually between 8 to 20.

If a baseline OUR is established for a system - presence of toxic susbstances in that system can be identi-

ﬁed by a drop in OUR. SOUR value above the optimal range indicates an increasing F:M - Higher BOD

load with less than adequate microorganisms (represented by MLVSS) indicating a young sludge and the

potential for straggler ﬂoc. A lower than optimal SOUR value indicates insufﬁcient food to support the

microorganisms indicating an older sludge and the potential for pin ﬂoc.

11.2.5 Food to Mass ratio (F:M)

In order to establish the amount of the MLSS to be maintained in the aeration tank, the parameter Food

to Mass Ratio (F:M) is used. F:M is the ratio of the "food" – the mass of primary efﬂuent BOD entering

the aeration basin to the mass of the microorganisms in the aeration basin —MLVSS. Common ranges for

F:M for a conventional activated sludge plant are from 0.15 to 0.5

11.2.6 Sludge age

• Sludge age is the average time (days) that a sludge particle (cell) spends in the system.

•

Sludge age dictates the presence and predominance of the different types of organisms and is

measured as Mean Cell Retention Time (MCRT) or Solids Retention Time (SRT).

• Sludge age is the average time that a sludge particle (cell) spends in the system

• Conventional AS process - MCRTs are in the range of 5 to 15 days

• The desired MCRT for a given plant must be found experimentally

•

F:M and MCRT are inversely related: that is a long MCRT means a low F:M and a short MCRT

means a high F:M

•

Both, F:M and MCRT values vary from plant to plant and are established through a trial and error

process.

• The MCRT is calculated as:

MCRT(days) =

Total MLSS lbs in the aeration system (aeration tank

+ clarifier)

Total amount in lbs/day of suspended solids leaving the system (E f fluent SS+WAS solids)

MLSS in aeration tank (lbs)+MLSS in clarifier (lbs)

MCRT(days) =

E f fluent suspended solids (lbs/day)+ WAS SS (lbs/day)

11.2.7 Sludge settleability

•

A settleability test is conducted on the mixed liquor from the aeration reactor to measure the com-

paction and settleability characteristics of the secondary solids.

In the settleability test, a 1-liter mixed liquor sample is allowed to settle for 30-minutes in a gradu-

ated cylinder − settleometer.

The Sludge Volume Index (SVI) of the sludge is calculated using results from the 30-minute

settleability test and the MLSS value.

SVI is expressed in ml/g and it is essentially the volume (ml) of 1 gram of the MLSS after 30

minutes of settling.

Settled sludge volume in ml/l after 30 min

• SVI (ml/g)=

∗1000^mg

MLSS mg/l

•

SVI provides a more accurate picture of the sludge settling characteristics than settleability or

166

Chapter 11. Activated Sludge

MLSS alone.

• 50 to 120 ml/g SVI value is considered optimal.

• Higher SVI values indicate sludge that is slow to settle and not compacting well.

• When SVI values are approaching 200 ml/g, activated sludge process is considered to be "bulking".

•

Regular monitoring of SVI help identify changes occurring in the activated sludge process prevent-

ing settling problems before they occur.

Like F:M and MCRT, the optimal SVI value for each plant varies and is also established by trial

and error.

11.3 Microbiology

Bacteria are fundamental microorganisms in the stabilization of organic wastes and therefore of basic

importance in the secondary wastewater treatment process.

Bacteria (singular, bacterium) are single celled organisms and represent the simplest forms of life. These

organisms utilize soluble food and are capable of self-reproduction. Bacteria can be classiﬁed in many

different ways.

For wastewater bacteria are classiﬁed based on nutritive requirement, bacteria are classiﬁed as het-

erotrophic or autotrophic bacteria, although several species may function both.

1. Heterotrophic bacteria use carbon based organic compounds for both energy and food source. A

term commonly used instead of heterotroph is “saprophyte” which refers to an organism that lives

on dead or decaying organic matter. The heterotrophic bacteria are grouped into three classiﬁca-

tions, depending upon their action toward free oxygen.

a. Aerobes : Require free dissolved oxygen to live and multiply.

b. Anaerobes : Oxidize organic matter in the complete absence of dissolved oxygen.

c. Facultative : Use free dissolved oxygen when available but can also respire and multiply in

its absence, e.g. Escheichia coli.

2. Autotrophic bacteria use carbon dioxide as a carbon source and oxidize inorganic compounds for

energy. Autotrophs of greatest signiﬁcance in wastewater treatment are the nitrifying and sulfur

bacteria.

a. Nitrifying bacteria : These bacteria are responsible for the oxidation of ammonia in wastewa-

ter to nitrates.

b. Sulfur bacteria : These bacteria are involved in the formation of sulfuric acid from the dis-

solved sulﬁdes in the sewer conveyance systems.

The activated sludge is a complex ecosystem of aerobic microorganisms predominantly heterotrophic

bacteria along with some autotrophic bacteria and other trophic organisms including protozoas, rotifers and

nematodes which feed on the bacteria and particulate matter.

Protozoa are single-celled microscopic organisms, several hundred times larger than bacteria. It is the

protozoa we observe under a microscope since bacteria are actually too small to see. There are four types of

protozoa commonly found in activated sludge systems. They are identiﬁed by their method of movement

within the wastewater environment. The four types are:

11.3 Microbiology

167

Amoeba

StalkedCilliate

Flagellate

Suctorian

Other commonly found organisms included the multi-celled organisms - Metazoans such as Rotifers and

Nematodes

Rotifer

Nematode

•

The essence of the activated sludge system is to develop and and maintain a diverse population of

microorganisms (activated sludge) that treats wastewater and which can be managed.

The activated sludge should be such that it not only effectively remove the BOD, but also produces a

168

Chapter 11. Activated Sludge

ﬂoc that settles and compacts well in the secondary clariﬁer thereby producing a secondary efﬂuent

with a low BOD content.

•

The ﬂoc produced has a porous structure and is essentially an aggregate of bacteria and extracelu-

lar material including organic polymers - polysaccharides and other substances secreted by the

microorganisms, and remnants of dead microorganisms.

Typically, the trophic organisms such as the protozoas and rotifers which feed on the bacteria and

other particulates are generally present on the outer surface of the ﬂocs.

• A good quality activated sludge is brown in color with some froth and a slight musty odor.

•

The settleability of the sludge is measured as SVI which is the volume of settled sludge in milliliters

occupied by 1 gram of dry sludge solids after 30 minutes of settling in a 1,000 mL graduated cylin-

der or a settleometer.

• SVI of a good settling sludge is typically between 50 - 150 ml/g.

• MCRT is the key factor which establishes the organism mix in the activated sludge.

• As MCRT increases, the size and complexity of the organisms increases.

•

The types of organisms predominating in the mixed liquor give an indication as to the age of the

sludge.

1. At low MCRT of about less than 4 days the simpler protozoas - amoebae and ﬂagellates

dominate.

2. With an increase in MCRT more complex organisms such as the free swimming ciliates and

stalked ciliates appear.

3. At even higher MCRT, metazoans such as the rotifers and nematodes may be found

11.4 Activated sludge process control

•

To ensure optimal treatment in the activated sludge process - maximum BOD removal with a well

settling mixed liquor, controlling the inventory of solids (biomass) is a key process control parame-

ter.

•

New solids (biomass) are produced in the activated sludge reactors due to the consumption of the

BOD by the microorganisms.

• The solids faction of the mixed liquor which separate (settle) in the secondary clariﬁer are

–

Either returned to the inﬂuent end of the activated sludge reactor as Return Activated Sludge

(RAS), or

– removed from the system as Waste Activated Sludge (WAS).

•

Thus, RAS and WAS pumping controls are used for maintaining the desired amount of solids in the

system.

11.4.1 Controlling waste activated sludge (WAS) pumping rate

For process stability and control the wasting should be continuous and not intermittent and any increase

or decrease in wasting should be made gradually, i.e., 20 - 25 percent per day. One of the following ap-

proaches is utilized to establish the WAS ﬂow control

Constant Mean Cell Residence Time (MCRT)

Typical MCRT for conventional AS process ranges from 5 to 15 days and 20 to 30 days for extended

aeration and for a given plant, the desired MCRT is established based upon operational experience. WAS

ﬂow control can be established to maintain the desired MCRT range for a given plant.

11.4 Activated sludge process control

169

Constant food:mass (F/M)

This parameter is based upon the ratio of food (inﬂuent BOD) fed to the microorganisms each day to the

mass of microorganisms held under aeration. WAS ﬂow control can be used for controlling the mass of

microorganisms in the system to maintain the desired F/M ratio.

Constant Mixed Liquor Suspended Solids (MLSS)

A MLSS concentration which provides the best efﬂuent and highest removal efﬁciencies is selected and

WAS ﬂow control is established to maintain the desired MLSS concentration level. Wasting is increased

when the MLSS concentration is higher and decreased or stopped when the MLSS concentration is lower

than the desired level.

11.4.2 Controlling the return activated sludge (RAS) rate

RAS ﬂow rate affects the concentration and the sludge age. Return Activated Sludge Control... To prop-

erly operate the activated sludge process, For conventional activated sludge operations, the RAS ﬂow is

generally about 30 to 100% of the incoming wastewater ﬂow. Changes in the activated sludge quality will

require different RAS ﬂow rates due to settling characteristics of the sludge. The following approaches are

used to control the RAS ﬂow rate:

Constant RAS ﬂow rate control - RAS pumping independent of inﬂuent ﬂow

•

Results in a continuously varying MLSS concentration that will be at a minimum during peak

inﬂuent ﬂows and a maximum during minimum inﬂuent ﬂows. This occurs because the MLSS are

ﬂowing into the clariﬁer at a lower rate during peak ﬂow when being removed at a constant rate.

Similarly, at minimum inﬂuent ﬂow rates, the MLSS are being returned to the aeration tank at a

higher rate than are ﬂowing into the clariﬁer

• Maximum solids loading on the clariﬁer occurs at the initial start of peak ﬂow periods.

•

The clariﬁer will have has a constantly changing depth of sludge blanket as the MLSS moves from

the aeration tank to the clariﬁer and vice versa.

This RAS control option is especially advantageous for smaller plants as it entails little effort/minimal

automation for RAS pumping control.

A disadvantage of using the constant ﬂow approach is that the F/M is constantly changing. The

range of F/M ﬂuctuation due to the effect of short term variation in the MLSS (because of hydraulic

loading) is generally small enough so that no signiﬁcant problems arise due to using this approach.

Constant percentage RAS ﬂow rate control

•

This approach requires a programmed method for maintaining a constant percentage of the aeration

tank inﬂuent wastewater ﬂow rate. The program may consist of an automatic ﬂow measurement

device, a programmed system, or frequent manual adjustments.

•

The variations in the MLSS concentration due to the diurnal ﬂow proﬁle are reduced and the F/M

varies less.

The MLSS will remain in the clariﬁer for shorter time periods, which may reduce the possibility of

denitriﬁcation in the clariﬁer.

Clariﬁer is subjected to maximum solids loading when the clariﬁer contains the maximum amount

of sludge. This may result in solids washout with the efﬂuent.

170

Chapter 11. Activated Sludge

Establishing RAS ﬂow rate based on a solids balance approach

The following equation can be used for controlling RAS ﬂow rate to maintain a target MLSS concentra-

tion given the inﬂuent ﬂow and the RAS solids content.

Applying a mass balance approach to the secondary clariﬁer:

Mass of solids in = Mass of solids Out

(Q+Q_RAS)∗MLSS = Q_RAS∗SS_RAS+Q_WAS∗SS_RAS

Notes:

• The solids leaving with the efﬂuent - Q∗SS_EFFis negligible

• Also, SS_WAS= SS_RAS

Rearranging:

Q∗MLSS−Q_WAS∗SS_RAS

SS_RAS−MLSS

=⇒ Q_RAS

As Q_WAS∗SS_RAS« Q∗MLSS and can be neglected

Rearranging:

Q∗MLSS

Q_RAS

SS_RAS−MLSS

11.5 Process trouble shooting

The most common activated sludge quality issues include:

1. Straggler ﬂoc: This condition shows small, light, and ﬂuffy ﬂoc particles with poor settling char-

acteristics. It occurs due to young sludge or low MCRT/MLSS levels and may be related to an

inadequate microorganism population or an excessive BOD load (high F:M) which causes a log

growth situation. The cells become dispersed rather than ﬂocculated, settleability is poor, and the

efﬂuent becomes turbid. In this condition oxygen is used up quickly due to the high metabolism

rate, and sludge production is high. One tell-tale sign of this condition is the production of huge

amounts of a billowing white foam.

2. Pin ﬂoc (Dispersed Growth): Here ﬂocs - pin ﬂoc is seen in the efﬂuent. These are larger dimension

spherical particles. The sludge settles well in the settleability test but the supernatant is cloudy.

This ﬂoc consist mostly of ﬂoc-forming bacteria without a ﬁlament backbone. This condition

occurs at starvation condition when all of the inﬂuent BOD has been used up (low F:M) and the old

sludge age organisms are now in endogenous respiration.

3. Rising sludge: Under this condition, sludge settles well, compacts on the bottom of the clariﬁer,

then starts to rise in clumps and patches to the surface. Rising sludge is typically evidenced as

brown in color. Rising sludge occurs due to either denitriﬁcation or sludge septicity due to an

excessive detention time in the clariﬁer.

4. Bulking and foaming: Typical bacteria present in activated sludge include spherical, rod shaped

11.5 Process trouble shooting

171

and ﬁlamentous. The presence of ﬁlamentous bacteria in activated sludge provide the important

structural element for the bacterial ﬂoc. Bulking is indicated by SVI > 150 ml/g. However, certain

condition cause an excessive growth of ﬁlaments which interferes with the settling (bulking) and

cause foaming during aeration. Identifying which ﬁlaments are dominating in the system through

a microscopic evaluation will help the operator to understand the condition in the treatment system

so that corrective changes can be made.

Filaments proliferation

a. Bulking: Activated sludge bulking is marked by poor compaction – high SVI₃₀and poor solids-

liquid separation evidenced by high efﬂuent TSS. To resolve bulking issue, it is important to ﬁnd

its cause and take remedial actions. This is accomplished through a microscopic examination

to identify the type of the ﬁlament to provide a clue on the cause. Short term remedy to bulking

includes Chlorinating – 2-3 mg/l per 1000 lbs MLVSS or use of polymers.

b. Foaming: Activated sludge foaming is caused mostly by two ﬁlaments: Nocardia spp. and Mi-

crothrix parvicella (there are other non-ﬁlament causes of foaming - including excessive presence

of surfactants. Both of these ﬁlaments have three causes in combination: (1) high grease and oil;

(2) longer sludge age; and (3) low oxygen conditions or septicity. Foaming if left uncontrolled

could potentially pose a permit compliance issue if the foam spills into the efﬂuent. Additionally,

Nocardia from the secondary sludge could cause digester foaming.

Filament Type

Cause

Remedy

Low aeration basin DO concentration for the applied F/M leads to ﬁlamen-

tous bulking by several ﬁlaments. Typically a minimum DO of 2 mg/l is

required for F:M upto 0.5. The condition may also be remedies by raising the

MLSS concentration (decreasing the F/M)

Low D.O. ( For the ap-

plied organic loading )

S.natans, Type 1701 and H.hydrossis

Increase the F:M by reducing the MLSS concentration. When reduction in

the MLSS concentration may not be feasible due to impact on nitriﬁcation,

other suitable methods such as the use of a selector may be considered

M.parvicella, Nocardia species, Type 0041,

Type 0675, Type 1851 and Type 0803

Low Organic Loading

Rate ( Low F:M )

Thiothrix I and II, Beggiatoa species, N.

limicola II, Type 021N, Type 0092, Type

0914, Type 0581, Type 0961 and Type 0411

Septic Wastes / Sulﬁdes (

High organic acids )

Reduce septicity using pre-aeration and through use of chemicals which

prevent septicity and reduce sulﬁdes

Nutrient Deﬁciency ( N

and / or P ) ( Industrial

Wastes Only )

Thiothrix I and II and Type 021N (N defe-

ciency) N. limicola III (P defeciency)

Nocardia species, M.parvicella and Type

1863

High Grease / Oil

The aeration basin pH should be maintained in the range 6.5 to 8.5. Low pH,

<6.5, may cause the growth of fungi and fungal bulking. The aeration basin

pH can be adjusted using caustic, lime or magnesium hydroxide.

Low pH ( Below pH 6.0 )

/ Oil

Fungi

Table 11.1: Troubleshooting Activated sludge ﬁlaments

11.6 Process modiﬁcations

173

11.6 Process modiﬁcations

11.6.1 Conventional activated sludge process

Process description

The primary efﬂuent (AS inﬂuent) - Q, along with the return activated sludge (RAS) - Q_RAS, are intro-

duced at the head of the aeration tank. The oxygen demand is the highest at this point and decreases

uniformly as the wastewater travels from one end of the tank to the other.

Figure 11.1: Conventional Activated Sludge Process

Process advantages and applications

The conventional activated sludge process is most suitable for low-strength, domestic wastes with minimal

peak load considerations as this process is susceptible to failure from shock loads. Shock Load is elevated

strength (higher BOD and contaminant levels) in wastewater for brief period of time.

Reasons for modifying the Conventional Activated Sludge Process

Several factors contribute to the need for modifying the Conventional Activated Sludge Process:

•

Unique characteristics of the inﬂuent ﬂow which may include - high BOD loads, potential toxic

loads, storm ﬂows, loads with high carbonaceous BOD but low nitrogen

• Space constraints

• Energy and labor (O&M) constraints

• Constraints requiring low solids production

11.6.2 Conventional step feed process

Process description

In this process modiﬁcation, the primary efﬂuent is fed to the aeration tank at several different locations

within the tank. The distribution of the inﬂuent along the length of the aeration reactor allows for distri-

bution of the load (BOD) over the aeration tank and reduce oxygen sags in the aeration tank. The return

sludge is introduced separately from the primary efﬂuent and in many cases is allowed a short re-aeration

period before mixing with primary efﬂuent.

Process advantages and applications

•

Less aeration volume required to treat the same volume of wastewater compared to the conven-

tional aeration.

• Better control in handling shock loads

• There is potential for:

174

Chapter 11. Activated Sludge

Figure 11.2: Conventional step feed

– lower applied solids to the secondary clariﬁer

– lower oxygen requirement

•

The step-feed process can be utilized on a variable basis using various modes (varying percent of

the inﬂuent at the feed points) depending on the inﬂuent situation and the selection of the proper

mode depends on the inﬂuent characteristics and the capabilities of the plant to handle the charac-

teristics

Figure 11.3: Step feed modiﬁcation for Biological Nutrient Removal (BNR)

11.6.3 Contact stabilization

Process description

• This process modiﬁcation requires two aeration tanks.

1. Return Sludge Re-aeration tank:

–

This tank is for separate re-aeration of the return activated sludge before it is mixed with

the incoming wastewater

– The residence time in the stabilization tank is typically 4 to 6 hours

2. Mixed Liquor Aeration tank (or Contact tank):

– This tank is for mixing of the primary efﬂuent with the re-aerated return activated sludge

–

Here, the organisms quickly take in and store large amounts of food from the primary

efﬂuent.

– The residence time in this tank is approximately 30 to 90 minutes

Process advantages and applications

Because the contact stabilization system has a reservoir of microbes in the re-aeration tank, it is better

able to withstand solids washouts that accompany major storms and avoid microbial kill-offs related to

short-term toxic dumps

11.6 Process modiﬁcations

175

Figure 11.4: Contact Stabilization

11.6.4 Pure oxygen system

Process description

Very similar to conventional activated sludge except that high purity oxygen is supplied to aeration tank

(reactor) rather than air and this process treats more wastewater in a smaller space in a shorter period of

time. The Aeration tanks (reactors) are enclosed and sealed to prevent the oxygen from escaping from

the tanks. Pure oxygen from an oxygen generation system or is introduced in the headspace of the reactor.

Mechanical surface aerators agitate the mixed liquor (essentially splash the liquid in the headspace) to

dissolve the oxygen into the mixed liquor. High DO levels are maintained in the aeration tanks (4 mg/l to

15 mg/l). Roughly 90%-99% gaseous O₂goes in and 45-70% gaseous O₂is ventilated at the efﬂuent end

of the reactor.

Figure 11.5: Pure Oxygen System

Process advantages and applications

• Pure oxygen systems are generally more stable and can treat large volumes of wastewater.

• It treats more wastewater in a smaller space in a shorter period of time

176

Chapter 11. Activated Sludge

11.6.5 Sequencing batch reactors

Process description

Rather than treating a continuous ﬂow through a number of tanks, Sequencing Batch Reactors (SBRs)

treat ﬂows in “batches” through a single tank - "reactor" through sequencing stages. It performs the same

treatment steps as a conventional AS plant, but they do it in sequenced stages rather than in a continuous

ﬂow stream. Most commonly preliminary treated wastewater is directly fed to the SBR.

Figure 11.6: Sequencing Batch Reactor

Typically SBR is operated in ﬁve stages:

1. Filling: Inﬂuent raw wastewater enters the reactor

2. Mixing & Aeration: This stage is where the BOD consumption and biomass production occurs.

Mixing and aeration may be alternated to accomplish nitriﬁcation (during aeration) and denitriﬁca-

tion (during mixing - without aeration - anoxic).

3. Settling - This is the clariﬁcation stage. - to allow for the activated sludge ﬂoc to settle.

4. Extraction - in this stage the supernatant treated wastewater is removed (for disposal) and part of

the settled sludge is wasted.

5. Idle - This stage is essentially to prepare the reactor and the biomass to receive the raw wastewater

- Filling stage. In this stage some aeration/mixing may be provided. The biomass in the reactor

enters the endogenous phase as the food is absent. The biomass metabolizes the stored food and the

organisms feed on one another.

11.6 Process modiﬁcations

177

Process advantages and applications

• SBRs are ideal for small facilities as the process can be easily automated

•

SBRs are very stable due to the high sludge ages maintained and the fact that all treatment takes

place in a single tank

•

SBR construction is less costly due to the elimination of secondary clariﬁers and sometimes diges-

tion facilities (fewer structures in general)

11.6.6 Extended aeration

Figure 11.7: Extended Aeration

Process description

This process typically directs raw sewage ﬂow (without grit/screenings) into the aeration tank where it

undergoes treatment for an extended period (usually 8 hours or more – 15 hours typical) followed by

clariﬁcation. Mixed liquor concentrations are usually maintained at high concentrations. Typical F:M

ratios are about 0.05 to 0.15 and MCRT of about 30 days.

Process advantages and applications

•

This process modiﬁcation is appropriate for small facilities that have light solids loadings – plants

typically treating less than 1 MGD – as the aeration tank volume requirement is larger

In addition to consuming all the incoming food, the microbes consume all food stored within cells -

endogenous respiration, and other dead microbes in the system

•

• Extended aeration does not produce as much waste sludge as other process modiﬁcations

•

The oxidation ditch is a variation of the extended aeration process. In the oxidation ditch, the aer-

ation basin is ring or oval shaped. Aeration rotors or brushes cause the wastewater to ﬂow around

the ring and also aerate the wastewater.

178

Chapter 11. Activated Sludge

Figure 11.8: Extended Aeration - Oxidation Ditch

11.6.7 Krauss process

Process description

The process involves use of two aeration tanks. One aeration tank (Aeration Tank B in the graphic), is

for blending return activated sludge (RAS) with anaerobic digester supernatant or sludge and aerating

the solution. The other tank receives primary efﬂuent (Aeration Tank A), return activated sludge, and the

aerated mixture of return sludge and digester supernatant/sludge. Air (oxygen) is added to both aeration

tanks.

Figure 11.9: Krauss Process

Process advantages and applications

This process modiﬁcation is widely used when wastewater contains a high ratio of carbonaceous to ni-

trogenous material in the wastewater (nitrogen deﬁcient). This typically occurs when there is signiﬁcant

contribution of wastewater from canneries or dairies. When the little nitrogen present in the inﬂuent

wastewater is consumed, further consumption of carbonaceous material ceases. The digester supernatant

or sludge being typically rich in nitrogenous material provides the required nitrogen for the nitrogen

deﬁcient process stream.

Chapter Assessment

1. What is the activated sludge ﬂoc made of

2. What properties of activated sludge ﬂoc are key to the effectiveness of the activated sludge process

3. List the activated sludge process control parameters (names/description)

4. List the key design differences between a rectangular and circular primary clariﬁer

5. Why is it important to ensure having a "good" microbiological composition of the activated sludge

process (5 points)

6. What is/are the main factor/s that control the microbial population

7. MLVSS represents the

fraction of the MLSS

8. Optimal range of SVI is between

9. Straggler ﬂoc is associated with [type1] sludge while pin-ﬂoc is associated with

Bulking and foaming is due to bacteria

sludge

10. Portion of the activated sludge ﬂoc settled in the clariﬁer that is returned to the front of the aeration

basin to seed the incoming primary efﬂuent is called [name]

11. Which one of the following statements is TRUE regarding the various modiﬁcations of the acti-

vated sludge process

a. MCRT of 5 to 10 days is typical for extended aeration

*b. Typical hydraulic detention times in the contact tank of the contact stabilization process need

only be 0.5 to 1.0 hour.

c. F to M ratios of 0.03 to .1 are appropriate for the step-aeration mode of the activated sludge

process

d. Pure oxygen activated sludge ﬂoc often has a large population of rotifers.

e. Step feed -aeration involves decreasing the air being fed along the length of the aeration tank.

12. What is the signiﬁcance/importance of measuring OUR and SOUR and what are their respective

units of measurement

Correct Answer(s):

13. List the advantages and disadvantages of the constant RAS ﬂow control

180

Chapter 11. Activated Sludge

Correct Answer(s):

14. Name and describe the two RAS control approaches

Answer the following related to activated sludge ﬂoc:

15. What is the activated sludge ﬂoc made of (3 points)

16. Activated sludge is an anaerobic process

a. True

*b. False

17. Secondary treatment is mainly to remove the organic content of the wastewater

*a. True

b. False

18. The contents of an aeration tank utilized in activated sludge treatment is referred to as mixed liquor.

*a. True

b. False

19. In conventional activated sludge plants, six to eight hours of aeration detention time is used for

acceptable plant operation.

*a. True

b. False

20. Bulking occurs in primary clariﬁers and is associated with improper scum removal.

a. True

*b. False

21. Contact stabilization is a modiﬁcation of the conventional activated sludge system.

*a. True

b. False

22. In secondary settling tanks, the sludge pumping considerations would be the same as in primary

settling tanks.

a. True

*b. False

23. The main function of a launder in a secondary clariﬁer is to prevent scum and other ﬂoatables from

leaving with the efﬂuent ﬂow

a. T

@Incorrect. Launder collects and conveys the efﬂuent ﬂow. Efﬂuent bafﬂes prevent scum and other

ﬂoatables from leaving with the efﬂuent ﬂow

*b. F

@Correct. Launder collects and conveys the efﬂuent ﬂow. Efﬂuent bafﬂes prevent scum and other

ﬂoatables from leaving with the efﬂuent ﬂow

24. Excessive ﬁlamentous bacteria in activated sludge is typically controlled by bleach addition to RAS

*a. True

b. False

25. SVI is a measure of the sludge volume that needs to be wasted

a. True

*b. False

26. pH has little effect on the activated sludge plant

a. True

*b. False

27. Bulking is caused by excessive ﬁlamentous bacteria

11.6 Process modiﬁcations

181

*a. True

b. False

28. Extended aeration involves operating the activated sludge process at a high F: M ratio

a. True

*b. False

29. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will

occur.

*a. True

b. False

30. In the activated sludge process, the wastewater oxygen demand may be separated into two cate-

gories: carbonaceous and nitrogenous

*a. True

b. False

31. MCRT refers to the average number of days that a “cell” remains in an activated sludge system.

*a. True

b. False

32. In activated sludge treatment a young sludge age is marked by a low F:M ratio

a. True

*b. False

33. The "M" in the F:M ratio is the mass of mixed liquor suspended solids in the aeration basin

a. True

*b. False

34. The SVI test is used for establishing amount of sludge to be wasted

a. True

*b. False

35. A WAS or RAS ﬂow change of 25% in one day will have little impact on the activated sludge

treatment process

a. True

*b. False

36. Rotifers are the dominant microorganisms in a young activated sludge

a. True

*b. False

37. The F in the F to M ratio refers to the pounds of mixed liquor volatile suspended solids under

aeration in an activated sludge plant.

a. True

*b. False

38. The use of F:M ratio for controlling the activated sludge process implies the need for higher mass

of microorganisms to treat a stream with a higher BOD

*a. True

b. False

39. The white billowing foam commonly seen during the startup of the activated sludge plant is caused

by low F:M ratio

a. True

*b. False

40. When an activated sludge plant is ﬁrst started, one should expect to see foaming

182

Chapter 11. Activated Sludge

*a. True

b. False

41. MLVSS represents the

42. Optimal range of SVI is between

43. Straggler ﬂoc is associated with [

sludge

fraction of the MLSS

sludge while pin-ﬂoc is associated with

bacteria

44. Bulking and foaming is due to

45. Portion of the activated sludge ﬂoc settled in the clariﬁer that is returned to the front of the aeration

basin to seed the incoming primary efﬂuent is called [name]

46. The basic objective in the activated sludge process is to maintain balanced conditions in the aera-

tion basin, this balance is called:

a. Endogenous respiration

*b. Food/microorganism ratio

c. Equilibrium status

d. Mass balance ratio

47. In the activated sludge treatment process, there are several control methods. One method is to

maintain a BOD:MLVSS ratio. This is commonly referred to as:

a. MCRT.

b. SA.

c. SA:SDI.

*d. F:M.

e. TS:SRT

48. In calculating the detention time in an aeration tank, which one factor would not be considered?

a. tank volume

b. RAS ﬂow

c. plant ﬂow

*d. MLSS concentration

e. none of the above

49. The BOD loading rate divided by the quantity of microorganisms present in the biological reactors

(aeration tanks) is known as:

a. organic loading

b. toxicity

c. hydraulic loading

*d. food to microorganism ration F:M

50. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what

type of activated sludge process modiﬁcation?

*a. Extended aeration

b. Conventional

c. Complete mix

d. Oxidation ditch

51. Two major operational difﬁculties which sometimes occur in activated sludge secondary clariﬁers

are:

*a. Low D.O. and algae growth

b. Short circuiting and scum accumulation

c. Rising sludge and bulking sludge

11.6 Process modiﬁcations

183

d. Long detention time and short MCRT.

52. A thick, scummy, dark tan foam on the surface of an activated sludge aeration tank is an indication

of:

*a. Aeration tank is underloaded (high MLSS.

b. Aeration tank is overloaded (low MLSS.

c. Excess grease in raw wastewater

d. Excess phosphates (detergents. in raw wastewater

53. A good quality of activated sludge is shown by:

a. Black color and very small particle size

b. Finely dispersed milky white particles

c. A chocolate brown MLSS that does not settle well in the jar test

d. A sludge that settles in one minute in the jar test

*e. A chocolate color which settles out in 20-30 minutes with a D.O. of 2.0

54. An aerobic treatment process is one that requires the presence of:

a. Ozone

b. organic oxygen

c. no oxygen

d. combined oxygen

*e. dissolved oxygen

55. An increasing F/M ratio and decreasing MCRT indicates

*a. Excessive solids wasting causing a decrease in solids inventory

b. Inadequate solids wasting causing an increase in the solids inventory

c. Decreased hydraulic load increasing the sludge detention time

d. Operation is normal

56. A rapid and signiﬁcant increase in ﬁlamentous organisms in the mixed liquor may be expected to:

a. Result in a far better efﬂuent because of the great amount of surface area for absorption

b. Plug up the return sludge pumps because the ﬁlaments hang upon valves and gaskets in the

sludge line

c. Lead to much denser return sludge because the ﬁlaments would tend to strain the dispersed cells

of ordinary organisms out of the efﬂuent

*d. Cause bulking of the sludge solids to the point that some solids might be swept out along with

an otherwise clear liquid phase and result in turbid, poor quality efﬂuent

e. Lead to a much lower F/M ratio because the ﬁlaments are so totally insoluble.

57. In the activated sludge treatment process, there are many control methods. One method is to main-

tain a constant BODs:MLVSS ratio. sludge treatment process. This is commonly referred to as:

a. MCRT

b. SA

c. SA: SDI

*d. F:M

e. TS:SRT

58. The SVI of activated sludge is deﬁned as:

a. the volume of settled mixed liquor after 30 minutes or settling·

b. the weight in grams of 200 ml of settled activated sludge

*c. the volume in ml of 1 gram of activated sludge after 30 minutes of settling

d. the total volume of MLSS in the aeration tank

184

Chapter 11. Activated Sludge

e. the volume of settled sludge in the secondary clariﬁer

59. The amount of air required in the operation of an activated sludge aeration tank is independent of

the:

a. temperature

b. ﬂow

c. detention time

d. organic loading

*e. none of the above

60. The successful operation of an activated sludge plant requires the maintenance of proper solids

concentration in the system. One major limiting factor is:

a. mixed liquor tank volume

b. efﬂuent ﬂow

*c. air supply

d. chlorine demand

e. none of the above

61. The main difference between primary and secondary clariﬁers is the:

a. overall dimensions

b. type of outlet weirs

*c. density of sludge

d. detention period

e. ﬂow distribution

62. Given the data below, what is the most likely cause of the extended aeration facility problem?

DATA: DO level high

Blower normal

Wastewater characteristics normal

Drop pipe air control valves open

Surface turbulence high

a. Air relief valve stuck shut

*b. Blower speed too fast

c. Blower speed too slow

d. Drop pipe air control valves not open far enough

63. What test is used to determine the organic matter found in the mixed liquor?

a. COD

b. MLSS

*c. MLVSS

d. TOC

64. Fixed porous plate diffusers can be cleaned by scrubbing with

a. Detergent

b. A strong acid solution

*c. A strong chlorine solution

d. A weak sodium hydroxide solution

65. A 30 minute settleability test MLSS sample should be collected:

a. At the primary clariﬁer efﬂuent

b. In the return sludge line

c. Where the return sludge mixes with the aeration basin contents

11.6 Process modiﬁcations

185

d. At the aeration basin inﬂuent

*e. At the aeration basin outlet

66. A consulting engineer has recommended addition of a roughing ﬁlter and intermediate clariﬁer

between your primary clariﬁer and aeration basin to better handle increasing industrial loads. This

addition would:

a. Be the best form of ﬂow equalization available

b. Remove most of the ﬁxed dissolved solids

c. Reduce drastically the ﬁne dissolved matter

d. Cost a lot and do nothing

*e. Reduce the organic load on the aeration basin

67. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what

type of activated sludge process modiﬁcation?

*a. Extended aeration

b. Conventional

c. Complete mix

d. Oxidation ditch

68. Two major operational difﬁculties which sometimes occur in activated sludge secondary clariﬁers

are:

*a. Low D.O. and algae growth

b. Short circuiting and scum accumulation

c. Rising sludge and bulking sludge

d. Long detention time and short MCRT.

69. Two major operational difﬁculties which sometimes occur in activated sludge secondary clariﬁers

are:

*a. Low D.O. and algae growth

b. Short circuiting and scum accumulation

c. Rising sludge and bulking sludge

d. Long detention time and short MCRT.

70. Possible techniques for controlling ﬁlamentous organisms in an activated sludge process include:

*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite

b. Lower DO levels in aeration bans so ﬁlamentous organisms cannot breathe or respire

c. Lower F/M level to starve ﬁlamentous organisms

d. Stop wasting to allow activated sludge bugs to gain control

71. Possible techniques for controlling ﬁlamentous organisms in an activated sludge process include:

*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite

b. Lower DO levels in aeration bans so ﬁlamentous organisms cannot breathe or respire

c. Lower F/M level to starve ﬁlamentous organisms

d. Stop wasting to allow activated sludge bugs to gain control

72. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what

type of activated sludge process modiﬁcation?

*a. Extended aeration

b. Conventional

c. Complete mix

d. Oxidation ditch

73. An aerobic treatment process is one that requires the presence of:

186

Chapter 11. Activated Sludge

a. Ozone

b. organic oxygen

c. no oxygen

d. combined oxygen

*e. dissolved oxygen

74. An increasing F/M ratio and decreasing MCRT indicates

*a. Excessive solids wasting causing a decrease in solids inventory

b. Inadequate solids wasting causing an increase in the solids inventory

c. Decreased hydraulic load increasing the sludge detention time

d. Operation is normal

75. A rapid and signiﬁcant increase in ﬁlamentous organisms in the mixed liquor may be expected to:

a. Result in a far better efﬂuent because of the great amount of surface area for absorption

b. Plug up the return sludge pumps because the ﬁlaments hang upon valves and gaskets in the

sludge line

c. Lead to much denser return sludge because the ﬁlaments would tend to strain the dispersed cells

of ordinary organisms out of the efﬂuent

*d. Cause bulking of the sludge solids to the point that some solids might be swept out along with

an otherwise clear liquid phase and result in turbid, poor quality efﬂuent

e. Lead to a much lower F/M ratio because the ﬁlaments are so totally insoluble.

76. During severe cold weather operation of an activated sludge plant biological activity and clariﬁer

sludge settling is reduced. White of the following might help?

*a. Increase the MLSS

b. Decrease the MLSS

c. Increase the D.0.

d. Decrease the D.0.

e. Add ammonia

77. Excess white foam in an aeration basin can be corrected by

a. Decreasing the aeration rate

b. Decreasing detention time

*c. Increasing the MLSS

d. Decreasing the MLSS

e. Increasing aeration rate

78. Given the following data, what is the most likely cause of the activated sludge problem?

DATA: The aeration tanks in an activated sludge plant have maintained a stable white foam with a

brownish tint less than one inch thick. ·

BOD removals have been at their normal high efﬁciency.

Settling of the activated sludge in the secondary clariﬁers has been good - as is normal.

Air supplied to the system has been a normal 30,000 cfm, with a consistent DO of 2.5 mg/L.

MLSS has been maintained at 2,500 mg/L - normal.

Gradually during your shift the DO has risen to 5.0 mg/L.

a. A toxic substance has affected the activated sludge.

b. BOD loading on the aeration system has increased.

c. Increased BOD loading has caused a corresponding increase in activated sludge activity.

*d. No change.

79. Given the following data, what is the most likely cause of the secondary sedimentation tank prob-

11.6 Process modiﬁcations

187

lem?

DATA: Sludge depth in tank too high.

Tank efﬂuent turbid.

Tank efﬂuent requiring above normal chlorine dosage. Sweeparms in tank bottom operating.

Return activated sludge ﬂow to aeration tank low.

Controls on return activated sludge pump on automatic.

Control sensors for return sludge operating normally.

a. Accuracy of sludge depth measurement.

b. Return activated sludge pump worn, needing repair.

*c. Speed of sweeparms travel.

d. Sweep arm overload tripped.

80. How many gallons of paint will be required to paint the walls of a 40 ft long x 65 ft wide x 20 ft

high tank if the paint coverage is 150 sq. ft per gallon. Note: We are painting walls only. Disregard

the ﬂoor and roof areas.

*a. 28 gallons

b. 63 gallons

c. 35 gallons

d. 56 gallons

81. If there is an insufﬁcient supply of air or oxygen being introduced into the aeration tank of an

extended aeration plant, the liquid in the tank will likely

a. Contain a very ﬁne light brown ﬂoc.

b. Contain very small air bubbles.

*c. Have a black or blackish appearance and an offensive odor.

d. Have a dishwater appearance and a greasy odor.

82. If the return sludge pump does not function the effect on other unit processes will be to:

a. Tum the aeration basin inﬂuent dark

b. Increase chlorine residual

*c. Increase efﬂuent suspended solids

d. All the above

e. None of the above.

83. If the sludge depth in a secondary sedimentation tank is too high, what will happen?

a. Decreased turbidity in efﬂuent.

b. Return activated sludge will have lower oxygen demand.

c. Settleable solids from aeration tank will increase.

*d. Sludge may become septic.

84. If you must waste sludge from an activated sludge plant the maximum rate is:

*a. 20 % per day

b. 40 % per day

c. 60 % per day

d. 75% per day

e. 100 % perday

85. In an activated sludge system, what is perhaps the most important parameter affecting biological

activity?

a. pH.

b. Alkalinity.

188

Chapter 11. Activated Sludge

*c. Dissolved oxygen.

d. Temperature.

86. Mean cell residence time (MCRT. represents the theoretical time that a microorganism stays in the

activated sludge system. The typical values for most activated sludge processes are:

a. 3 - 30 days

*b. 3 - 15 days

c. 5 - 15 days

d. 5 - 20 days

87. Nocardia is associated with a particular type of brown, viscous scum or foam on the surface of

the activated sludge aeration tank. One operational strategy that has been somewhat successful in

reducing the severity of this foam is:

a. to increase the plant’s MCRT.

*b. to decrease the plant’s mixed liquor concentration.

c. to operate at an F:M ratio of less than 0.025.

d. to spray the foam with ﬁne mist water sprays.

e. to increase the luxury DO concentration at the end of the aeration tank.

88. One limitation in using constant mixed liquor volatile suspended solids (MLVSS) or mixed liquor

total suspended solids (MLTSS) as the control methodology for activated sludge treatment is

*a. In practice it is not possible to operate at a constant MLTSS or MLVSS.

b. It is based on consistency of raw waste load which seldom exists.

c. Most facilities don’t have the lab equipment necessary to determine MLVSS.

d. None of the above.

89. Deﬁne Nocardia & problems associated with it. List 5 methods of controlling Nocardia.

Response:

a Deﬁne Nocardia & problems associated with it.

•

Nocardia is a type of ﬁlamentous organism which on overgrowth in the mixed liquor is

the cause of foaming during aeration in the activated sludge process. Nocardia prolifer-

ation is associated with the following three causes in combination: (1) high grease and

oil; (2) longer sludge age; and (3) low oxygen conditions or septicity.

Nocardia develops a persistent, viscous brown foam scum. If not skimmed properly,

this can cause an increase in both suspended solids and BOD and cause scum spill over

on to the catwalks and make them sticky and slimy. Additionally, Nocardia from the

secondary sludge could cause digester foaming.

b List 5 methods of controlling Nocardia.

i. Reduce sludge age by increasing wasting rate. This would also effectively reduce MLSS

concentration in the aeration tank and increase F:M.

ii. Mechanically skim and remove the foam.

iii. Enforce industrial waste control program and manage the collections cleaning proce-

dures to minimize the amount of grease and oil in the inﬂuent wastewater.

iv. Chlorinate the RAS stream or the mixed liquor return or by spraying directly into the

aeration tank.

v. Use of mannich polymer to enhance settling and removal of the Nocardia ﬁlaments

11.6 Process modiﬁcations

189

90. Clearly show how SOUR & OUR are different. List two possible causes for a sudden 40% decrease

in OUR during the last 12 hours of operation. What lab test might conﬁrm or refute the two causes

of this sudden decrease in OUR. Response:

a Clearly show how SOUR & OUR are different.

•

Oxygen Uptake Rate (OUR) involves measurement of the amount of oxygen used up by

the microorganisms in the mixed liquor using a DO probe and is expressed in unit time

of mg/L-hr (ppm O₂consumed per hour). By knowing the OUR, we can establish the

activity of the microorganisms in the aeration tank and know if they are consuming the

oxygen provided for removing organic matter. For conventional activated sludge process

the typical OUR values range from 10 to 30 mg/L-hr

Speciﬁc Oxygen Uptake Rate (SOUR) provides the OUR information based on the

concentration of microorganisms present. SOUR is obtained by dividing OUR with

MLVSS.

l−hr

The value is indicated and measured in terms of unit of

∗1000

MLVSS

Optimal range of SOUR is usually between 8 to 20.

List two possible causes for a sudden 40% decrease in OUR during the last 12 hours of

operation.

i. Presence of toxic substance in the mixed liquor inhibiting the normal biological activity

in the aeration basin

ii. Excessive loss of biomass due to inadvertent wasting

c What lab test might conﬁrm or refute the two causes of this sudden decrease in OUR.

i. Run a MLSS test to see if adequate biomass is present

ii. Run a COD test to ensure adequate amount of organics (food) is present

(If there is sufﬁcient biomass and food present, the only reason why the OUR would

have declined would be because of toxic substance entering the system)

91. The control and calculation of RAS ﬂow rates are important considerations in the Operation of

activated sludge wastewater treatment plants. Most commonly the operator Either sets the return

rate (Qr) at a constant ﬂow or as a constant percentage of ﬂow.

(a) Identify two effects that each of these approaches will have on plant operations.

Assume that a normal diurnal variation in both wastewater ﬂow and strength.

(b) The so-called “solids balance approach” (shown below) may be used to Mathematically esti-

mate Qr. What two assumptions are made in deriving this formula?

[Qr + Q] x MLSS Conc. = Qr x RAS Conc.

comment on these values:

Q. ... . . . . . . . . . . . . . . . . . ..2.0 MGD

MLSS. . . . . . ..... . . . . 2350 mg/L

RAS.. . . . . . . . . . . .. . . .7350 mg/L

SV30.. . . . . . . . . . . . . . . . . 320 mLs/g

12. Nutrient Removal

12.1 Importance of removing nutrients

•

Plant nutrients - nitrogen and phosphorous, if present in wastewater efﬂuent discharge promote

growth of plant and algal matter in the receiving waters causing destruction of the normal aquatic

life mainly due to oxygen depletion - eutrophication.

Because of the potential impacts of the presence of these nutrients in wastewater efﬂuent on the re-

ceiving waters, limits on the levels of these nutrients is typically stipulated in the treatment plant’s

wastewater discharge permit.

Typically, conventional secondary treatment processes are designed primarily remove the organics

from the wastewater. Secondary treatment process designed to additionally remove nutrients is

deemed as tertiary or advanced treatment is termed as Biological Nutrient Removal (BNR).

12.2 Nitrogen Removal

Nitrogen exists in many forms and changes from one form to another in the environment as part of the

nitrogen cycle below.

•

In raw domestic wastewater, nitrogen exists primarily as organic nitrogen (40%) and ammonia

nitrogen (60%).

The sum of organic nitrogen and ammonia nitrogen is referred to as “Total Kjeldahl Nitrogen”

(TKN).

−

The important forms of nitrogen as part of wastewater treatment are: N₂(Nitrogen Gas), NO₂

(Nitrite), NO₃⁻(Nitrate), NH₃(Ammonia), NH₄⁺(Ammonium Ion), and Organic Nitrogen.

Typical concentrations of the nitrogen constituents in wastewater are tabulated below.

Ammonia (NH₃) and ammonium (NH⁺₄) are both commonly referred to as “Ammonia-nitrogen (NH₃-N)”

These two forms of nitrogen can rapidly change from one to the other depending on pH and temperature.

192

Chapter 12. Nutrient Removal

Figure 12.1: Nitrogen Cycle

The ﬁgure below illustrates the relative distribution of ammonia (unionized ammonia) and the ammonium

ion (ionized ammonia)in water as a function of pH.

•

In biological treatment processes, the organic nitrogen is quickly changed to ammonia nitrogen by

natural processes.

•

Ammonia-nitrogen is toxic to aquatic life and toxicity is affected by both, temperature and pH of

water.

– Toxicity of ammonia nitrogen increases as temperature increases

–

Ammonia nitrogen is toxic when present as unionized NH₃. This unionized ammonia NH₃

is the predominant form of ammonia nitrogen at higher pH. The ammonia nitrogen (NH₃-N)

toxicity is lower when present as ionized ammonia (NH⁺₄)

–

Juvenile ﬁsh is more susceptible to ammonia toxicity. Therefore, ammonia toxicity effects

will may be more pronounced during the ﬁsh breeding season in spring

12.2.1 Nitrogen Removal Methods

Nitriﬁcation and denitriﬁcation

This is the most common ammonia removal process and is accomplished during secondary treatment. It is

a two-step process:

1. Nitriﬁcation This involves conversion of ammonia to nitrate

2. Denitriﬁcation This involves conversion of nitrate to gaseous nitrogen which escapes from the

waster.

NITRIFICATION

Nitriﬁcation is a two step process:

Step 1: Nitrosomonas and other similar bacteria oxidizes ammonium to nitrite via hydroxylamine.

2NH₄⁺+O₂→ 2NH₂OH +2H⁺

12.2 Nitrogen Removal

193

NITROGEN IN WASTEWATER

Forms of Nitrogen

Formula

Found in

Typical

Concentration

Ammonia/Ammonium

NH₃/NH₄

Inﬂuent wastewa-

ter

30-50 mg/l

30-60 mg/l

1-40 mg/l

Total Kjeldahl Nitrogen

(Ammonia/Ammonium + Organic Nitrogen)

TKN

TIN

Wastewater

efﬂuent

Total Inorganic Nitrogen

(Ammonia/Ammonium + Nitrite + Nitrate)

Wastewater

efﬂuent

−

Nitrate

NO₃₋

Nitriﬁed efﬂuent

1-35 mg/l

0.1-2 mg/l

NO₂

Partially nitriﬁed

efﬂuent

Table 12.1: Forms of Nitrogen in Wastewater

Figure 12.2: Ammonia-Ammonium Equilibrium

194

Chapter 12. Nutrient Removal

NH₄⁺+1.5O₂→ NO₂⁻+2H⁺+H₂O

Step 2: Conversion of nitrite to nitrate by nitrobacter and other similar bacteria

−

NO₂⁻+0.5O₂→ NO₃

•

The above two reactions are generally coupled and precede rapidly to the nitrate form; therefore,

nitrite levels at any given time are usually below 0.5 mg/L.

Nitriﬁers (responsible for the consumption of nBOD) compete with the organotrophs which con-

sume the cBOD in the secondary reactor.

Nitriﬁcation occurs at the tail end of the secondary treatment reactor when the cBOD is low and the

organotrophs are not as active

• Nitriﬁcation occurs 3-4 times slower than carbonaceous oxidation

•

For nitrosomonas to generate 1 part of new cells it needs 30 parts of ammonium ions and nitrobac-

ter needs 100 parts of nitrite ions to generate 1 part new cells. Due to the high demand of the nitro-

gen based ions, the reproductive rates of nitriﬁers are very low thus plant upsets can take nitriﬁers

weeks to recover as opposed to days or hours for carbon bacteria.

•

For each 1-gram of NH₃-N oxidized to NO₃, 0.15 grams of new bacteria cells are formed. Sludge

generation from nitriﬁers is minimal in a wastewater treatment plant.

Controlling Factors for Nitriﬁcation:

1. Dissolved oxygen Oxygen levels are critical for nitriﬁcation. Higher levels of oxygen are required

for nitriﬁcation than for carbon removal. While 1 part of O₂is needed for removing 1 part of

cBOD, 4.5 parts of O₂are needed for every part of NH₄⁺N (nBOD) to be degraded. In order

for nitriﬁcation to occur, dissolved oxygen levels of 1.0 to 4.0 mg/L are usually maintained in

the aeration tanks. Maximum nitriﬁcation occurs at DO levels of about 3.0 mg/L. Nitriﬁcation

will therefore result in signiﬁcantly higher power cost due to the needs for maintaining the higher

oxygen level

2. Alkalinity and pH

Nitriﬁcation leads to a depletion of alkalinity in the activated sludge mixed liquor. Nitro-

somonas and Nitrobacter use carbonate as their carbon (food) source and the ammonia is just their

energy transfer source. As the total alkalinity of wastewater is due to the individual contributions

of carbonate, bicarbonate and hydroxide ions, consumption of carbonate and bicarbonate ions by

the nitriﬁers results in the depletion of alkalinity. Alkalinity provides a buffer for pH change. If

the alkalinity is reduced beyond certain levels, further formation of acidic metabolic byproducts of

bacterial activity may lead the pH to decline inhibiting bacterial activity. 7.14 parts of alkalinity

are required for each part of ammonia removed. Nitriﬁcation rates are rapidly depressed as

the pH is reduced below 7.0. pH levels of 7.5 to 8.5 are considered optimal. That is why alkalinity

is extremely critical in nitriﬁcation. An alkalinity of 60 mg/L in the secondary treatment reactor

(aeration tank, trickling ﬁlter, RBC, etc.) is generally required to ensure adequate buffering

3. MCRT, F/M, or Sludge Age For nitriﬁcation to occur the activated sludge treatment process

needs to be operated at higher MCRT as the reproductive rates of nitrﬁers is low. An MCRT

of greater than 8 days is typically considered essential for nitriﬁcation to occur.

4. Wastewater temperature Nitriﬁcation is inhibited at lower wastewater temperatures in wastew-

ater treatment plants. To achieve the same level of nitriﬁcation, longer detention time may be

needed in the winter versus the summer months since the activity drops signiﬁcantly. Lower tem-

perature effects on Nitriﬁcation may be partially mitigated by increasing MLVSS and MCRT. The

12.2 Nitrogen Removal

195

Figure 12.3: Weather Effects on MCRT Requirements for Nitriﬁcation

optimal temperature range for nitriﬁcation is between 60^◦to 95^◦degrees F. Below 40^◦nitriﬁcation

will probably not occur.

5. Inhibition to nitriﬁcation by toxic compounds Many compounds can be toxic to nitriﬁers- Nitro-

somonas and Nitrobacter. These include unionized ammonia, heavy metals, solvents and cyanide.

196

Chapter 12. Nutrient Removal

DENITRIFICATION

•

Dentriﬁcation is a microbiological process in which the nitrate formed during nitriﬁcation is re-

duced by bacteria in anoxic conditions to molecular nitrogen which is insoluble in water and es-

capes into the atmosphere.

•

A large percentage of the bacteria in the activated sludge process are capable of denitriﬁcation.

In the absence of free molecular oxygen – under anoxic conditions, these bacteria present in the

activated sludge process utilize the nitrate and nitrite to degrade the cBOD.

NO⁻₃+cBOD → NO₂⁻+CO₂+H₂O

NO⁻₂+cBOD → N₂+CO₂+H₂O

• Approximately 50% of the alkalinity lost during nitriﬁcation is recovered during denitriﬁcation.

•

The quantity of cBOD available is critical for denitriﬁcation to occur. Complete denitriﬁcation

usually occurs at a cBOD to nitrate and nitrite ratio of approximately 3:1. Given the need for the

presence of adequate cBOD for the denitriﬁcation step, a cBOD supplement (a suitable organic

compound - example: alcohol) is utilized in certain conﬁgurations.

•

If the aeration reactor efﬂuent is nitriﬁed, denitriﬁcation can occur in clariﬁer. The absence of

aeration in the clariﬁer creates anoxic conditions which makes the condition favorable for denitri-

ﬁcation to occur. Denitriﬁcation in the secondary clariﬁer is not desirable as the nitrogen evolved

will cause the sludge settling in the clariﬁer to rise to the surface.

Three of the more common methods to denitrify the nitriﬁed efﬂuent are:

Method 1.The aeration tank is followed by a denitrifying tank which is not aerated but is supplied with

organic substrate (cBOD source) such as methanol or acetic acid.

Figure 12.4: Denitriﬁcation Method 1:

Organic substrate addition

Method 2.The mixed liquor along with RAS is recirculated to an anoxic zone at the inlet of the aera-

tion tank where it is mixed with the inﬂuent. The anoxic zone has mixing provided but no aeration. The

inﬂuent ﬂow provides the necessary cBOD for the denitriﬁcation process.

12.2 Nitrogen Removal

197

Figure 12.5: Denitriﬁcation Method 2:

Mixed liquor recirculation

Method 3.The aeration tank consists of alternating oxic (aerated) and anoxic zones

Figure 12.6: Denitriﬁcation Method 3:

Alternating oxic-anoxic zones

Breakpoint chlorination

• Breakpoint chlorination is a means of eliminating ammonia using chlorine

•

As chlorine is added to water containing ammonia, it converts ammonia into chloramines which in

the presence of additional free chlorine forms nitrogen gas which is released to the atmosphere

Chlorination of a water containing ammonia results in the following:

• an initial increase in combined chlorine residual

• followed by a decrease in the combined chlorine residual along with ammonia concentrations

•

followed by an increase in free chlorine residual and near complete removal of ammonia as nitro-

gen gas.

• Chlorine reacts with ammonia to form chloramines

1. First the free chlorine in contact with ammonia forms monochloramine and water

– Monochloramine has disinfection properties

– Dominates when Cl:N mass ratio is 0 to 5:1

– The breakpoint curve rises at about 1:1 during monochloramine formation

NH₃+HOCl → NH₂Cl(monochloramine)+H2O

198

Chapter 12. Nutrient Removal

2. Monochloramine reacts further with chlorine to give dichloramine and water

HOCl +NH₂Cl → NHCl₂+H₂O

Also, monochloramine auto decomposes into dichloramine

2NH₂Cl → NHCl₂+NH₃

– Between dichloramine is formed between 5:1 and 7:1 Cl:N mass ratio

– When you are getting signiﬁcant dichloramine, the breakpoint curve will start dropping

NH₂Cl +HOCl → NHCl₂(dichloramine)

at pH

> 7.5, monochloramine is the dominant chloramine species as pH decreases from 7.5,

dichloramine becomes the dominant chloramine species increases in the chlorine to nitrogen

dose ratio results in corresponding increases of nitrogen trichloride, but only when the pH is

< 7.4

3. Formation of nitrogen trichloride from the reaction of chlorine and dichloramine does not

typically occur as it is the favored product at low pH - <4

NHCl₂+HOCl → NCl₃(nitrogentrichloride)

4. Additional free chlorine +chloramines → H⁺+H2O+N₂

•

Chloramines have disinfection properties albeit much lower than free chlorine ( 5% of free avail-

able chlorine) but last much longer in the system than free chlorine.

After the breakpoint, free chlorine residuals develop. Free chlorine residuals usually destroy odors,

kill microorganisms and oxidize organic matter.

Breakpoint chlorination is the application of sufﬁcient chlorine beyond the chlorine demand to

maintain a free available chlorine residual.

Theoretically chlorine requirement = Wt. NH₃-N x 7.6

in practice (Margin of safety) = Wt. NH₃-N x 10

•

Thus, breakpoint chlorination is possible if ratio of Cl₂to ammonia exceeds 10:1 then free Cl₂

may exist. Cl₂demand will be high because the reaction of free Cl₂with nitrite and other organic

compounds.

Theoretically, while microorganisms are killed as the chlorine demand is being satisﬁed, disin-

fection is generally the result of chlorine residual or the amount of chlorine remaining after the

chlorine demand has been satisﬁed.

The following chart is a graphical explaination of the concept of Breakpoint Chlorination

12.2 Nitrogen Removal

199

• Point A is at the beginning of chlorine application

•

Between Points A and B, the chlorine dosage produces no residual because of an immediate chlo-

rine demand caused by fast-reacting ions from metal salts and H₂S.

• Point B is the beginning of the reaction between chlorine and ammonia present

• Mono and dichloramines are formed between points B and C

•

Zone 1 - between points A and C, is the combined zone and has mono and di chloramines and

ammonia. Mono chloramines is a stable disinfectant while dichloramines is a strong disinfectant

but unstable.

•

After the maximum combined residual is reached (point C), further chlorine doses decrease the

residual due to chloramine oxidation to dichloramine, occurring between points C and D. This is

Zone 2 - Breakpoint Zone

•

Point D represents the breakpoint - the point at which chlorine demand has been satisﬁed and

additional chlorine appears as free residuals

Between points D and E, free available residual chlorine increases in direct proportion to the

amount of chlorine applied. This is Zone 3 which is the free chlorine zone and has hypochlorous

acid but no ammonia.

•

Factors that affect breakpoint chlorination are initial ammonia nitrogen concentration, pH, tempera-

ture, and demand exerted by other inorganic and organic species

Weight ratio of chlorine applied to initial ammonia nitrogen must be 8:1 or greater for the break-

point to be reached. If the weight ratio is less than 8:1, there is insufﬁcient chlorine present to

oxidize the chlorinated nitrogen compounds initially formed

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Chapter 12. Nutrient Removal

•

When instantaneous chlorine residuals are required, the chlorine needed to provide free available

chlorine residuals may be 20 or more times the quantity of ammonia present. Reaction rates are

fastest at pH 7-8 and high temperatures

12.3 Phosphorous removal

•

Limits related to phosphorous concentrations are commonly found in wastewater treatment facili-

ties’ NPDES discharge permits.

•

Phosphorous removal as part of the treatmnent process accomplishes meeting the NPDES require-

ment also allows for its recovery for agricultural use.

• Phosphorous is very reactive and occurs in nature as phosphate (PO₄⁻).

•

In wastewater phosphorous is present mainly in an inorganic form as orthophosphate (typically

about 10 ppm) and as organic phosphate (typically 6 mg/l).

The inorganic phosphate is contributed by detergents and household cleaning products such as

soaps

Organic phosphate is contributed by human wastes and food residues as phosphate based molecules

play a key role in biological (plant and animals) energy and lifecycles and present in sugars, phos-

pholipids, and nucleotides. The organic phosphates are decomposed to orthophosphate.

12.3.1 Phosphorous removal methods

Chemical precipitation method

In the chemical based phosphorous removal method, aluminum and iron salts such as ferric/ferrous chlo-

ride and aluminum sulfate react with soluble phosphate to form solid precipitates that are removed by

solids separation processes including clariﬁcation and ﬁltration.

Typically, the phosphate precipitant is applied in both the primary clariﬁer feed and also just before the

secondary clariﬁer.

Figure 12.7: Chemical removal of phosphorous

Enhanced biological phosphorus removal (EBPR)

• For the removal of phosphorous as part of the activated sludge treatment process:

12.3 Phosphorous removal

201

–

The process is designed to promote the growth of speciﬁc aerobic microorganisms generally

known as phosphorus accumulating organisms (PAOs) the activated sludge process.

Examples of these bacteria in activated sludge: Acinetobacter, Rhodocyclus

These bacteria have the ability to store substantial quantities of intracellular (inside the

bacterial cell) phosphorus.

PAOs maybe upto 40% phosphorous by weight.

–

The process is operated by providing alternating anaerobic and aerobic conditions in the

aeration basin of the activated sludge treatment process.

EBPR process description:

•

In the anerobic part of the aeration basin, the aerobic bacteria produce VFAs by breaking down

organic matter in the inﬂuent wastewater. Thus, the readily biodegradable material - VFAs, is

amply available for the PAOs when in the anaerobic environment.

–

Here the PAOs consume the volatile fatty acids (VFAs) produced by the anaerobic fermen-

tation of the organic matter in the anaerobic zone. Note: This is a fermentation process not

anaerobic digestion (in digestion VFAs are destroyed and converted to methane).

– The VFAs are converted into energy rich, carbon based, storable food source.

–

The energy for this conversion of the VFA into storable food comes from the breakdown of

the PAOs polyphosphate reserves.

– Ortho phosphate is released during this phase.

–

After exposure to enough VFA, the PAOs energy reserves will be depleted and become

stressed.

•

In the following aerobic phase of the process the PAOs multiply and take up phosphate to replenish

the supplies depleted in the anaerobic phase.

–

By oxidizing the carbon reserves built up in the anaerobic phase, PAOs are able to store more

phosphate under aerobic conditions than was released under anaerobic conditions because

considerably more energy is produced by aerobic oxidation of the stored carbon compounds

than was used to store them under anaerobic conditions.

Eventhough the EPBR requires only a good aeration, its success in removing phosphorous is dependent on

the presence of adequate quantities of readily degradable organic material in the secondary inﬂuent.

Chapter Assessment

1. Nitriﬁcation is conversion of

2. Nitrosomonas are

Correct Answer(s):

a. Autotrophic

(type) bacteria

3. Denitriﬁcation requires

Correct Answer(s):

a. anoxic

condition

4. Presence of adequate quantity of

Correct Answer(s):

a. vfa

is critical for phosphorous removal

5. In nitriﬁcation,

Correct Answer(s):

a. 4.57

parts of oxygen required per part of ammonium nitrogen

6. In domestic water

Correct Answer(s):

a. 40

7. Nitriﬁcation is conversion of

8. In domestic water

Correct Answer(s):

a. 40

% of nitrogen is present as ammonia/ammonium

9. The approximate pH for good nitriﬁcation is closest to

a. 6.5

b. 7.0

*c. 8.5

d. 10.0

10. Which of the following biological processes can produce alkalinity?

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Chapter 12. Nutrient Removal

a. Carbonaceous BOD removal

*b. Denitriﬁcation

c. Nitriﬁcation

d. Phosphorus removal by chemical addition with ferrous chloride

11. In a typical biological nitriﬁcation system, an oxygen demand of

parts oxygen per part of

ammonia oxidized is exerted in the nitriﬁcation process.

a. 2.37

b. 5.47

c. 27

*d. 4.57

12. The biological nitriﬁcation process is carried out by bacteria that convert ammonia nitrogen to

nitrate nitrogen. the two (2) speciﬁc groups of bacteria that perform this conversion are:

*a. Nitrosomonas and nitrobacter

b. Flagellates and swimming ciliates

c. Flagellates and crawling ciliates

d. Crawling ciliates and swimming ciliates

13. Biological denitriﬁcation requires

a. Aerobic

conditions.

*b. Anoxic

c. Anaerobic

d. Facultative

14. Denitriﬁcation is an activated sludge plant involves:

a. Oxidation of ammonia to nitrate.

b. High concentrations of D.O. in the mixed liquor as it settle in the secondary clariﬁer.

c. Biological oxidation of nitrate to nitric oxide.

*d. Biological reduction of nitrate to nitrogen gas.

e. Poor compaction of mixed liquor as it settles in the secondary clariﬁer.

15. In a nitrifying activated sludge plant, it is important to maintain adequate D.O. in the mixed liquor

as it settles in the ﬁnal clariﬁer to prevent:

a. Death of organisms in the ﬂoc that “eat” organic materials.

*b. Denitriﬁcation.

c. Bulking.

d. Growth of ﬁlamentous organisms.

e. Death of ﬁlamentous organisms.

16. Nitriﬁcation in activated sludge does not involve:

a. The oxidation of ammonia to nitrite.

*b. Removal of oxygen from nitrate to form nitrogen gas.

c. The biological oxidation of nitrite to nitrate.

d. Higher concentration of D.O. in the aeration basin.

e. Control of alkalinity to stabilize pH.

17. Limits on the concentration total NH3 - N are frequently placed on secondary efﬂuents discharged

into sensitive receiving waters. An important consideration in the setting of these limits by the

RWQCB would be:

a. The concentration of algal cells in the receiving water.

*b. The pH & temperature of the receiving water.

12.3 Phosphorous removal

205

c. The salinity of the receiving water.

d. The population of nitrosomonas bacteria in the ﬁnal efﬂuent.

e. The concentration of copper ions in the ﬁnal efﬂuent because these ions form a non-toxic com-

plex with the ammonia nitrogen.

18. Many NPDES permits limit the amount of ammonia nitrogen that can be discharged into a stream

because:

a. Ammonia exerts an oxygen demand in the stream

b. Ammonia can be toxic to aquatic life

c. Ammonia reacts with chlorine which can interfere with disinfection

*d. Any of these

19. Possible techniques for controlling ﬁlamentous organisms in an activated sludge process include:

*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite

b. Lower DO levels in aeration bans so ﬁlamentous organisms cannot breathe or respire

c. Lower F/M level to starve ﬁlamentous organisms

d. Stop wasting to allow activated sludge bugs to gain control

20. Nitrogen and phosphorous are biologically signiﬁcant in waste water because:

a. Nitrogen is reduced to pure nitrogen gas when phosphorous is present as a catalyst and this leads

to nitrogen super saturation and rapid ﬁsh kills.

b. Nitrogen in the form of nitrates will reduce sodium thiosulfate and thus cause low results in the

D.O. test while phosphates will oxidize the iodine to iodide and cause high results.

*c. Nitrogen and phosphorous are essential nutrients that can enable aquatic plant growth to be-

come troublesome.

d. Nitrogen pentaphosphate is toxic to the organisms that metabolize.

21. What are the bacteria that remove ammonia in the activated sludge process

a. Filamentous bacteria

*b. Nitrosomonas

c. Stalked ciliates

d. E. Coli

22. The loading on a nitriﬁcation system consists of ammonia plus organic nitrogen and is typically

referred to as total nitrogen.

a. True

*b. False

23. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitriﬁed activated sludge

efﬂuent having a pH of 6.8 and a temperature of 67 deg. F. Ammonia nitrogen is found to range

from 2 to 3 mg/L in this efﬂuent. Disinfection in this efﬂuent might be difﬁcult because:

a. A temperature of 70 deg. F or higher is necessary in order to achieve effective disinfection.

b. Chloramines are present most of the time.

c. The chlorine dose rate is too low.

d. The pH of this efﬂuent will limit the effectiveness of free chlorine.

*e. The ratio of chlorine to ammonia nitrogen may make it difﬁcult at times to maintain adequate

chlorine residual.

24. During breakpoint chlorination:

a. A ratio of 5 parts of chlorine to 1 part of ammonia nitrogen is adequate to reach the breakpoint.

b. Ammonia nitrogen is reduced to nitrogen dioxide.

*c. As the breakpoint is approached, additional chlorine causes the chlorine residual to decrease.

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Chapter 12. Nutrient Removal

d. Signiﬁcant concentrations of dichloramine remain after the breakpoint is passed.

25. In the breakpoint chlorination method of ammonia nitrogen removal, parts of chlorine are required

for each part of ammonia removed.

a. 3.0

b. 6.7

*c. 7.6

d. 2.0

12.3 Phosphorous removal

207

26. The correct statement regarding the effect of nitrogen in chlorine disinfection is:

a. The reaction of ammonia with chlorine increases pH.

b. The reaction of ammonia with chlorine decreases pH.

c. The reaction of nitrite with free chlorine produces chloride.

*d. The reaction of ammonia with free chlorine produces chloramines.

e. The reaction of nitrate with free chlorine produces nitrogen gas.

27. You are the superintendent of a 100,000 gpd conventional activated sludge plant which discharges

into a shallow bay. Your NPDES permit currently sets a discharge limit of 5 mg/L for total ammonia-

nitrogen in your efﬂuent. However, the Regional Water Quality Control Board at the request of the

Department of Fish and Game will soon revise your plant’s discharge requirements and have even

lower total ammonia limits so that certain species of ﬁsh may be re-introduced into the bay. The

proposed maximum total ammonia-nitrogen concentration limit for November through March 15th

- 2.0 mg/L and 1.0 mg/L for the period March 16th through October 31st Answer the following

questions:

1. Limits on the concentration total NH₃- N are frequently placed on secondary efﬂuents dis-

charged into sensitive receiving waters. An important consideration in the setting of these

limits by the RWQCB would be:

a The concentration of algal cells in the receiving water.

b The pH & temperature of the receiving water.

c The salinity of the receiving water.

d The population of nitrosomonas bacteria in the ﬁnal efﬂuent.

The concentration of copper ions in the ﬁnal efﬂuent because these ions form a non-

toxic complex with the ammonia nitrogen.

2. Many NPDES permits limit the amount of ammonia nitrogen that can be discharged into a

stream because:

a Ammonia exerts an oxygen demand in the stream

b Ammonia can be toxic to aquatic life

c Ammonia reacts with chlorine which can interfere with disinfection

d Any of these

3. Impacts of new ammonia-nitrogen limits on plant operations would include:

a Process change to remove ammonia further: nitriﬁcation – denitriﬁcation

b Increased MCRT, higher activated sludge power demands and Lower sludge yields

c Potential impacts to disinfection strategy related to breakpoint chlorination

d Options [a] and [b]

e Options [a], [b] and [c]

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Chapter 12. Nutrient Removal

28. You are the operations supervisor of a wastewater treatment plant that consists of the following

unit processes: primary clariﬁers, activated sludge aeration tanks, secondary clariﬁers, a DAF

thickener used to thicken WAS, and anaerobic digesters. Anaerobic digester gas is used, when it

is available in sufﬁcient amounts, to fuel engines that drive blowers supplying air to the activated

sludge aeration tanks. Waste heat from these engines is sufﬁcient to heat the digesters. When

a sufﬁcient amount of gas is not available, then electric motors are used to drive blowers. The

regional water quality control board will soon impose a limit of 1.0 mg/L on the discharge of

ammonia-nitrogen. In order to meet this limit, your plant will be modiﬁed to achieve complete

nitriﬁcation.

Answer the following questions:

1. Demand Charge is measured in:

a). $/kWh

b). $/Hp

c). $/kW

d). $/day

2. Most signiﬁcant impact on energy consumption related to changes in mode of operation in

order to meet the new permit requirement will be due to:

a). Increased digester gas production

b). Additional sampling and testing requirements

c). Associated with meeting higher F:M requirements for nitriﬁcation

d). Higher aeration air requirements

3. Which of the following is true for Demand Charges:

a). Applies to residential customers also.

b). Based upon how much electricity used and the rate at which it is consumed

c). It is calculated based on 15-minute interval data

d). Options a) and c)

e). Options b) and c)

f). Options a), b) and c)

4. Digester gas production is expected to:

a). Increase.

b). Decrease

c). Remain the same

5. More power will need to be purchased because of:

a). Increased power demands related to denitriﬁcation

b). Meet power demand associated with higher F:M ratio requirements

c). Additional sludge pumping to digesters

d). Additional RAS pumping requirement

12.3 Phosphorous removal

209

29. Part A: You are the chief plant operator at a 3 MGD conventional activated sludge plant that dis-

charges into a nearby creek. Currently you operate this plant at an MCRT of between 4 to 5 days.

Your plant’s maximum daily total coliform limit is 23 MPN/100 ml. Chlorine disinfection to

achieve a total chlorine residual of 5 mg/L followed by dechlorination with sulfur dioxide had

been used to meet your efﬂuent total coﬁfom, limit. Recently, at the request of the state department

of ﬁsh and game, the RWQCB revised your plant’s NPDES pemiit in order to make the creek more

habitable for certain species of ﬁsh. Your plant will now be required to meet a ﬁnal efﬂuent total

ammonia-nitrogen (N-NH₃) limit of 2 mg/L during the winter months (November 1- March 1) and

1 mg/L during the rest of the year. Your plant’s total coliform limit was not revised.

Answer the following questions:

(a) Deﬁne total ammonia-nitrogen?

(b) Why are such low limits necessary? Why are there different limits in winter and summer?

operations

Part B: Prior to this revision of your NPDES permit, your plant rarely had a problem meeting its to-

tal colifom, limit. Within days of meeting the revised ammonia-nitrogen limit, you’ve experienced

frequent violations of your total coliform limit. Plant records show that maintaining a total chlorine

residual of 5 mg/L has been difﬁcult even though more chlorine has been used. Likewise, chlorine

use has increased signiﬁcantly. You’ve also noticed that at times the efﬂuent in the chlorine contact

chamber is "crystal clear" almost like a swimming pool. Plant records show no signiﬁcant changes

in plant ﬂow or inﬂuent characteristics (i.e. BOD, TSS, pH, etc.).

(a) How do these observations help explain why you are having total colifom, violations?

(b) Identify and brieﬂy discuss one step you might take to prevent these coliform violations.

Response:

Part A:

a Deﬁne total ammonia-nitrogen

•

Total ammonia nitrogen is the total amount of nitrogen in the forms of NH₃and NH⁺₄in

the wastewater.

b Why are such low limits necessary? Why are there different limits in winter and summer?

•

Presence of nitrogen in wastewater efﬂuent will promote plant & algae growth causing

eutrophication (oxygen depletion) of the water body in which the efﬂuent is discharge

impacting aquatic life

• Additionally, ammonia-nitrogen is toxic to aquatic life

•

Toxicity of ammonia is affected by temperature, ﬁsh breeding times and susceptibility of

juvenile ﬁshes to ammonia– thus the different limits during winter and summer

Identify and brieﬂy discuss three signiﬁcant impacts that these new limits will have on plant

operations

•

Process changes to remove ammonia further through implementation of nitriﬁcation –

denitriﬁcation as part of the activated sludge process

Potential impact to the disinfection process required to meet the coliform limit - break-

point chlorination

•

Associated impacts include:

– Increase MCRT – increased RAS pumping

– Lower sludge yields – less biosolids

– Additional oxygen requirements – more energy consumption – higher power costs

210

Chapter 12. Nutrient Removal

–

Potential need for alkalinity and cBOD supplements for facilitating nitriﬁca-

tion/denitriﬁcation

– Less digester gas production

Part B:

a How do these observations help explain why you are having total colifom, violations?

• Potential impact of nitrite accumulation on chlorine disinfection

• 5 ppm of chlorine used up for each part of nitrite present

• Chlorine being consumed by nitrite and is not available for disinfection

b Identify and brieﬂy discuss one step you might take to prevent these coliform violations.

• Measure nitrite levels at the efﬂuent end of the activated sludge reactors

• Increase DO for the nitriﬁcation step to facilitate the oxidation of nitrite to nitrate

12.3 Phosphorous removal

211

30. Deﬁne nitriﬁcation. Name four factors that effect nitriﬁcation and discuss ranges, requirements and

what lab tests are used.

Response:

a Deﬁne nitriﬁcation

•

In wastewater treatment, nitriﬁcation is used for the removal of ammonia nitrogen - the

predominant form of nitrogen in wastewater. Nitriﬁcation involves a two step biological

process where in the ﬁrst step ammonia NH₃is biologically oxidized by ammonia

oxidizing bacteria such as nitrosomonas to nitrite (NO₂), followed by the oxidation of

NO₂to nitrate (NO₃) by ntirite oxidizing bacteria such as nitrobacter.

Name four factors that effect nitriﬁcation and discuss ranges, requirements and what lab tests

are used.

i. Dissolved oxygen

4.5 parts of O₂are needed for every part of NH₄⁺N (nBOD) to be degraded. In order for

nitriﬁcation to occur, dissolved oxygen levels of 1.0 to 4.0 mg/L are usually maintained

in the aeration tanks. Maximum nitriﬁcation occurs at DO levels of about 3.0 mg/L.

Dissolved oxygen (DO)is measured using a DO probe.

ii. Alkalinity and pH

Presence of adequate alkalinity in the mixed liquor is critical as the nitriﬁcation pro-

cess consumes alkalinity. Alkalinity provides a buffer for pH change. If the alkalinity

is reduced beyond certain levels, further formation of acidic metabolic byproducts of

bacterial activity may lead the pH to decline inhibiting bacterial activity. 7.14 parts of

alkalinity are required for each part of ammonia removed. Nitriﬁcation rates are

rapidly depressed as the pH is reduced below 7.0. pH levels of 7.5 to 8.5 are consid-

ered optimal. An alkalinity of 60 mg/L in the secondary treatment reactor is generally

required to ensure adequate buffering.

Alkalinity is measured in the laboratory by titrating the sample with an acid until a

speciﬁc pH is reached. Alkalinity is reported as mg/l of calcium carbonate.

iii. MCRT, F/M, or Sludge Age

For nitriﬁcation to occur the activated sludge treatment process needs to be operated

at higher MCRT/sludge age (low F:M) as the reproductive rates of nitrﬁers is low. An

MCRT of greater than 8 days is typically considered essential for nitriﬁcation to occur.

iv. Wastewater temperature

Nitriﬁcation is inhibited at lower wastewater temperatures in wastewater treatment

plants. To achieve the same level of nitriﬁcation, longer detention time may be needed

in the winter versus the summer months since the activity drops signiﬁcantly. Lower

temperature effects on Nitriﬁcation may be partially mitigated by increasing MLVSS

and MCRT. The optimal temperature range for nitriﬁcation is between 60^◦to 95^◦de-

grees F. Below 40^◦nitriﬁcation will probably not occur.

v. Inhibition to nitriﬁcation by toxic compounds

Many compounds can be toxic to nitriﬁers- Nitrosomonas and Nitrobacter. These in-

clude unionized ammonia, heavy metals, solvents and cyanide.

212

Chapter 12. Nutrient Removal

31. What is meant by un-ionized ammonia and total ammonia nitrogen? Why is it important to set a

low limit on un-ionized ammonia-nitrogen?

Response:

a What is meant by un-ionized ammonia and total ammonia nitrogen?

Ammonia is the predominant form of nitrogen in wastewater. Ammonia (NH₃) can exist

as ammonia itself which is the unionized form or it could change to its ionized form - am-

monium (NH⁺₄) by absorbing a proton (H⁺). Total ammonia nitrogen is the sum of the con-

centrations of the unionized ammonia and the ammonium ions present. These two forms of

nitrogen can rapidly change from one to the other depending on pH and temperature.

b Why is it important to set a low limit on un-ionized ammonia-nitrogen?

i. Nitrogen in any of its forms, if present in wastewater efﬂuent discharge, promote growth

of plant and algal matter in the receiving waters causing destruction of the normal

aquatic life mainly due to oxygen depletion - eutrophication. Ammonia is sought by

nitrifying bacteria and is converted to nitrate at the expense of the oxygen present in the

water body.

ii. Additionally, ammonia specially in its unionized form is particularly toxic to aquatic life

12.3 Phosphorous removal

213

32. Deﬁne and explain importance of:

(a) Breakpoint chlorination

(b) Chlorine demand

(d) Chloromines

(e) Rotometer

Response:

a Breakpoint chlorination

When disinfecting with chlorine, as chlorine is added to wastewater, the presence of inorganic

and organic substances including ammonia in the wastewater, will exert a demand for chlo-

rine as chlorine is a strong oxidizing agent. This consumption of chlorine does not allow the

chlorine to be present as free chlorine - the strongest form of chlorine disinfectant. Break-

point chlorination is the point where the demand for chlorine has been fully satisﬁed and any

further addition of chlorine will show a proportional increase in free chlorine residual.

b Chlorine demand

The amount of chlorine used up as part of the reaction of chlorine with the inorganic and

organic substances present in wastewater is referred to as the chlorine demand.

c Chlorine residual

Chlorine residual is the sum of free chlorine and combined chlorine and it represents the

amount of chlorine available for disinfection.

d Chloramines

Chloramines which include monochloramine, dichloramine and trichloramine are products of

the reaction of chlorine with ammonia.

e Rotometer

Rotometer is a ﬂow measurement device most commonly used for measuring the ﬂow of

chlorine gas for disinfection.

13. Solids Treatment Overview

13.1 Why do we need to treat wastewater solids?

•

Sludge is generated from the wastewater treatment processes - settled solids and scum from pri-

mary and secondary treatment processes

• This sludge contain organic compounds and also elements that are beneﬁcial plant nutrients

• However, the organic solids in the sludge are not stable (i.e. they will decay) and include pathogens.

•

Prior to disposal, sludge has to be treated – stabilized, so that its disposal or reuse does not pose a

threat to public health.

Sludge treatment is very critical as it is an expensive process and sludge disposal is subject to strict

regulatory requirement.

Even solids are only a small component of wastewater, the solids treatment and disposal account

for a very substantial portion of wastewater treatment costs. Typically 40 to 60% of total wastewa-

ter treatment operations cost is attributable to sludge treatment and disposal.

NOTE: Solids removed during Preliminary Treatment, from barscreens and grit chambers are typi-

cally not treated as part of the solids treatment process. These solids are disposed off at a landﬁll

Typical solids treatment is comprised of the following three sequential steps:

1. Sludge thickening

2. Sludge stabilization

3. Sludge dewatering

13.2 Sludge thickening

Sludge thickening involves the removal of excess water from the primary and secondary sludge increasing

the solids content of the sludge and reducing the volume of sludge to be treated in the sludge stabilization

216

Chapter 13. Solids Treatment Overview

process. Sludge thickening reduces the volume of sludge that need to be handled in the sludge stabiliza-

tion step thereby reducing treatment cost.

•

There is an upper limit of the solids concentration that can be effectively treated (stabilized) as

increasing the solids concentration reduces its ability to be mixed and pumped easily. Typically the

sludge thickening process produces sludge with a solids content of less than 10%.

Beneﬁts of thickening to the sludge stabilization process include:

• Improved performance due to a lower volume of sludge

• Cost savings in the construction of new facilities

• Reduction in energy requirements as less water has to be heated

Typical methods used for sludge thickening include:

1. Gravity thickener - more suitable for primary sludge

2. Dissolved air ﬂoatation thickener - more suitable for lighter, ﬂufﬁer ﬂoc such as the secondary

sludge.

13.3 Sludge Stabilization

Sludge stabilization process produces solids that are deemed safe for eventual disposal. Federal Part 503

rule establishes requirements for the ﬁnal use or disposal of sewage sludge. The solids disposal methods

may include: land application, as a crop/vegetation fertilizer, placed on a surface disposal site for ﬁnal

disposal and ﬁred in an incinerator.

Biosolids is the term used for stabilized sludge which meets regulatory standards for beneﬁcial reuse

Sludge stabilization process results in the following:

1. Reduction in amount of solids

2. Pathogen reduction

3. Odor reduction

4. Reduction in vector attraction

The main processes involved in sludge stabilization include:

• Digestion - Aerobic or anaerobic

• Lime or alkaline stabilization

• Composting

• Long term storage in lagoons

• Thermal processes

• Incineration

13.4 Sludge Dewatering

Solids stabilized using digestion process has only a small percentage by weight of solids -less than 5%.

It therefore becomes necessary to dewater the stabilized sludge prior to hauling off-site for ﬁnal disposal.

Like thickening, the dewatering process does not treat the sludge. It increases the solids content to be-

tween 15 to 30 percent and the higher solids content of the stabilized sludge makes it easier to handle and

reduces costs associated with elements related to accomplishing the end objectives with the sludge – land

application, composting, drying, incineration or landﬁll.

Dewatering involves conditioning the sludge with a polymer and subjecting it to a physical process which

include:

1. Belt Filter Press

2. Centrifuge

14. Biosolids Regulations

14.1 Biosolids Deﬁnition

• Biosolids are treated solids produced as part of wastewater treatment

• Biosolids can be disposed or recycled beneﬁcially

•

Biosolids must meet standards established under Title 40 of the Code of Federal Regulations

(CFR), Part 503.

• Grit and screenings are not considered as biosolids

14.2 Biosolids Use/Disposal Methods

1. Land application

•

This is the most commonly used biosolids management method and is referred as a "recy-

cling" option.

This uses the organic and/or nutrient content of the biosolids to either condition and/or fertil-

ize crops or other vegetation grown in the soil

Land application allows for the beneﬁcial utilization of soil-enhancing constituents such as

plant nutrients and organic matter in the biosolids

2. Surface disposal which requires availability of a large land which is lined with an impermeable

material prior to the application of biosolids.

3. Incineration where the biosolids is burnt to ash. This method utilizes its organic content to limit the

amount of external fuel required to incinerate.

14.3 Biosolids Regulations

•

Part 503 rule applies to any person who applies biosolids to the land or ﬁres biosolids in a biosolids

incinerator, and to the owner/operator of a surface disposal site, or to any person who is a preparer

or generator of biosolids for use, incineration, or disposal.

218

Chapter 14. Biosolids Regulations

Table 5.1: Pollutant Concentration Limits and Loading Rates for Land Application

• Part 503 standard includes:

1. General requirements

2. Pollutant limits

3. Management practices

4. Operational standards, and

5. Requirements for the frequency of monitoring, record-keeping, and reporting

14.4 Title 40 CFR Part 503 Requirements

In order to land apply biosolids, each of the following three standards must be met.

14.4.1 Pollutant concentration limits

•

The biosolids produced must have concentrations of the 10 heavy metals listed, lower than the

Ceiling Concentration thresholds.

Sewage sludge exceeding the ceiling concentration limit for even one of the regulated pollutants is

not classiﬁed as biosolids and, hence, cannot be land applied.

Biosolids with heavy metal concentrations at or below the limit speciﬁed in the Pollution Concen-

tration thresholds are classiﬁed as "High Quality Biosolids"

Biosolids meeting pollutant concentration limits are subject to fewer requirements than biosolids

meeting ceiling concentration limits

14.4.2 Pathogen reduction

• Pathogens are disease causing organisms such as bacteria, viruses and parasites

14.4 Title 40 CFR Part 503 Requirements

219

•

The treatment method used for treating wastewater solids must meet standards related to pathogen

reduction

• Based upon the method used for treating wastewater solids, biosolids produced are classiﬁed as:

Class A Biosolids

•

The biosolids produced using the methods and standards identiﬁed are free of measurable pathogens.

There are no pathogen related site restrictions for the application of Class A biosolids - including

use in home gardens and landscaping

•

Processes to further reduce pathogens (PFRP) treatment, such as those involving high temperature,

high pH with alkaline addition, drying, and composting, or their equivalent are most commonly

used to demonstrate that biosolids meet Class A requirements

Class B Biosolids

•

For Class B biosolids, pathogens have been reduced to levels that are unlikely to cause a threat to

public health and the environment under speciﬁed use conditions.

Processes to signiﬁcantly reduce pathogens (PSRP), such as digestion, drying, heating, and high

pH, or their equivalent are most commonly used to demonstrate that biosolids meet Class B require-

ments

•

Site restrictions are imposed on the application of Class B biosolids to ensure minimizing the

potential for human and animal contact with the biosolids until environmental factors reduce the

pathogens to below detectable levels.

14.4.3 Vector attraction reduction

Vector reduction standards are for preventing transmission of pathogens via rodents, birds, and

insects from the land applied biosolids.

•

• The vector reduction rule requirements are based on the following two approaches:

1. Specifying organic matter decomposition processes viz., digestion and alkaline addition to

mitigate vector attraction

2. Biosolids sub-surface injection or incorporation within six hours so soil microbes out-

competes/eliminates pathogens

•

A minimum of 38% volatile solids reduction as part of the solids treatment process is a more common

method of demonstrating compliance with vector attraction reduction requirements

For biosolids to qualify for Exceptional Quality (EQ) Standards - the application of which is almost

unrestricted, it must meet all three of the following:

1. Pollutant concentration limits

2. Class A requirements

3. Vector attraction reduction standards

220

Chapter 14. Biosolids Regulations

Chapter Assessment

1. Grit and screenings from the preliminary treatment are also considered as biosolids

*a. True

b. False

2. A true statement regarding the term “biosolids” is:

a. The term is mandated for user by public law 92-500.

b. The term was developed by US EPA to deﬁne all biologically toxic precipitates.

*c. The term is recommended by WEF for “a primarily organic solids product, produced by

wastewater treatment processes, that can be beneﬁcially recycled”.

d. The term is used by the California Water Resources Control Board to include “all insoluble

matter derived from living aquatic organisms.

3. When you spread sludge on agricultural land, the annual application rate of cadmium in the sludge

should be less than 2 lbs/acre/year If your sludge contains 30 mg cadmium per kilogram of solids

and your plant produces 950,000 lbs per year of dry solids, how many acres do you need?

a. 3 acres

*b. 14 acres

c. 19 acres

d. 27 acres

222

4. Essay type question:

Chapter 14. Biosolids Regulations

(a) Deﬁne vector and pathogens.

(b) Class A & B landﬁll requirements: what you need to do to get your anaerobic digester to

produce class B sludge to prepare for landﬁll.

Response:

a Deﬁne vector and pathogens.

• Pathogens are disease causing organisms such as bacteria, viruses and parasites

•

Vectors are organisms such as rodents and insects that can carry disease by carrying and

transferring pathogens

Class A & B landﬁll requirements: what you need to do to get your anaerobic digester to

produce class B sludge to prepare for landﬁll.

For the digested sludge to qualify as Class B sludge it needs to meet the following related to

40 CFR Part 503:

• Meet the Minimum Concentration Standards of the Pollution Concentration Standards

• Meet the digestion sludge detention time-temperature requirements

• Meet the Vector Reduction Standards

c What unit processes can produce class A & B sludge?

For Class A sludge:

i. Composting

ii. Heat drying

iii. Heat treatment

iv. Thermophillic aerobic digestion

v. Beta ray irradiation

vi. Gamma ray radiation

vii. Pasteurization

14.4 Title 40 CFR Part 503 Requirements

223

For Class B sludge:

i. Aerobic digestion

ii. Air drying

iii. Anaerobic digestion

iv. Composting

v. Lime stabilization

15. Sludge Thickening

Prior to the sludge stabilization process such as anaerobic digestion, the solids content of the sludge is

increased by utilizing an appropriate thickening process.

Notes:

1) There is an upper limit of the solids concentration that can be effectively treated as increasing the solids

concentration reduces its ability to be mixed and pumped easily. Digesters can effectively treat sludges

with upto about maximum 7% - 9% solids.

2) If a 5000 mg/l (0.5%) sludge is thickened to 5% solids concentration - it will reduce the volume of

sludge by 90%

15.1 Advantages of sludge thickening

• Improved digester performance due to a lower volume of sludge

• Capital Cost savings associated with less digester volume requirements

• Operational costs savings - for sludge heating and mixing

15.2 Sludge thickening methods

1. Gravity thickener - more suitable for primary sludge

2. Dissolved air ﬂoatation thickener - more suitable for lighter, ﬂufﬁer ﬂoc such as the secondary

sludge.

15.3 Gravity thickener

The gravity thickener is designed and operated similar to a circular primary and secondary clariﬁer.

226

Chapter 15. Sludge Thickening

15.3.1 Principles of gravity thickener operations

•

Upon entering the tank from the center, the sludge solids settle under the inﬂuence of gravity and

these settled solids accumulate in the bottom of the gravity thickener as the sludge blanket. As

the sludge becomes thicker it helps squeeze out more water from the sludge increasing the sludge

solids content.

•

Typical solids loading rates of a gravity thickener used for thickening primary sludge is 20-30 lbs

TS/day-ft²

15.4 Dissolved air ﬂoatation thickener (DAFT)

227

•

A picket fence-like mechanism which is attached to the bottom sludge rake arms is primarily to

release entrapped gases. The sludge rake arms rotate at a very slow speed to ensure that it does

not cause turbulenece causing the sludge to rise. The sludge is gently raked towards a sump in the

center, from where the solids are withdrawn. The sludge rake encounter much higher torques than

the typical primary clariﬁer rake, and is therefore designed to be more stronger and heftier.

The thickened solids are drawn-off at regular intervals and the liquid fraction is decanted from the

top and returned to the primary clariﬁer. The typical sludge concentration factor is about 2.

•

• The sludge blanket is kept about three feet.

•

The sludge blanket depth is typically maintained so that the sludge does not turn septic and the

efﬂuent is relatively clear and free of solids.

Sludge Volume ratio (SVR) provides a means of regulating the detention time of sludge within the

blanket of the thickener

SVR is deﬁned as the volume of the sludge blanket divided by the daily volume Of sludge pumped

from the thickener.

SVR is a relative measure of the average detention time of solids in the thickener and is calculated

in days.

• Typical SVRs for primary sludge range from 0.5 - 2 days.

• Scum is removed from the top using a separate scum removal mechanism.

• Chemical additives may be utilized to enhance settling and control odors.

15.3.2 Elements of a gravity thickener

• center feed column and bafﬂe

• drive assembly

• scum removal system

• thickened sludge pump

• sludge rake with pickets

• efﬂuent weir

• sludge hopper and pump

15.3.3 Gravity thickener operational parameters

• type and quality of sludge

• hydraulic or surface loading rate (GPD/ ft²/day)

• solids loading rate (lbs/ ft²/day)

• sludge volume ratio - sludge detention time

• solids and hydraulic loading rates, and

• quantity and characteristics of the polymer used

15.4 Dissolved air ﬂoatation thickener (DAFT)

Opposite to the principle of the gravity thickener where the thickened sludge forms a blanket at the bot-

tom, in the DAFT the thickened sludge forms a blanket on the top surface. Primary sludge is not generally

suitable for thickening using DAFT as its solids are heavier and do not ﬂoat easily. The WAS from the

secondary clariﬁer which typically has a solids content of about 0.5% to 0.8% is thickened to about 4% to

6% concentration – thickened WAS (TWAS) using the DAFT.

228

Chapter 15. Sludge Thickening

15.4.1 Principles of DAFT operations

•

WAS is conditioned with cationic polymer and introduced into the DAFT. DAFTs typically operate

at a solids loading rate of 1-2 lbs TSS/hr-ft². Polymer feed ranges from 5-15 lbs of polymer/ton of

the WAS (feed) solids.

Recycled water from the DAFT is pressurized with air in the saturation tank and mixed with poly-

mer treated WAS as it is released at the bottom of the DAFT using the Back Pressure Control

Valve.

The dissolved air from the pressurized water is released as minute air bubbles rises upwards car-

rying with it the polymer ﬂocculated sludge to the surface. Adequate quantity of air is required to

ﬂoat the WAS solids. Air:Solids ratio is one of the key operating and control parameter. Typical

air:solids ratios in a DAFT are between 0.03 - 0.05 lb air per lb TS. PS: Density of air used for

air:solids ratio calculations - 0.075 lb air/ft³air - this value is given as part of the problem.

The thickened solids ﬂoating on the top are scrapped off the surface of the DAFT by ﬂights into

the TWAS sump from where it is pumped to the digesters. The subnatant - water below the solids,

part of it is used for the air pressurization in the saturation tank and the remaining is the underﬂow,

•

15.4 Dissolved air ﬂoatation thickener (DAFT)

229

which is returned back to the inﬂuent ﬂow.

•

The ﬂight speed is critical to the DAFT performance. Fast ﬂight speed would limit the thickness

and density of the sludge blanket while slower ﬂight speed would result in the thickened solids

layer getting more dense and thick. Excessively thick solids layer could result in the solids escap-

ing through the underﬂow.

15.4.2 Elements of a DAFT

• pressurization or saturation tank

• thickened sludge skimmer with drive assembly

• polymer dosing and injection system

• thickened sludge pump

• back pressure control valve

• underﬂow removal

• recycled ﬂow system

15.4.3 DAFT operational parameters

• saturation pressure

• solids loading rate (lbs TSS/hr- ft²)

• hydraulic loading rate (GPD/ ft²/day)

• feed solids concentration

• detention period

• air-to-solids ratio (lb air: lb solids)

• type and quality of sludge

• ﬂight speed

• solids and hydraulic loading rates, and

• quantity and characteristics of the polymer used (lbs polymer/dry ton solids)

Chapter Assessment

1. In a gravity thickener the depth of the sludge is kept minimal (<six inches) to avoid solids going

over the efﬂuent weir

a. True

*b. False

2. Sludge thickening is primarily conducted to reduce costs associated with biosolids hauling

a. True

*b. False

3. Gravity thickener is commonly used for sludge dewatering

a. True

*b. False

4. Sludge thickening is primarily conducted to reduce costs associated with biosolids hauling

a. True

*b. False

5. A DAF thickener has efﬂuent solids of 55 mg/L and ﬂoat solids of 2.0%. Solids loading and poly-

mer dosing is in the normal range. This data likely indicates:

a. This unit is operating normally

b. Too low air to solids ratio

c. Float blanket too thick

*d. Flight speed too fast

e. Flight speed too slow

6. An air ﬂotation thickener will produce a thin ﬂoat if:

*a. Flight speed too high and skimmer wiper not adjusted properly

b. Excessive air/solids ratio and polymer dosages too low

c. High dissolved oxygen and ﬂight speed too low

d. Polymer dosages too high and unit overloaded

7. An increase in the pool depth of a scroll-type centrifuge:

232

Chapter 15. Sludge Thickening

a. would not affect the moisture content of the cake.

b. would produce a drier cake

*c. would produce a wetter cake, but produce a greater solids recovery.

d. would not affect either solids recovery, nor cake moisture content.

e. would require an increase in the cationic polymer dosage.

8. A sludge thickened from 1% to 4% solids will be reduced in volume by how much?

a. no more than 4% of original volume

b. approximately 17% of original volume

*c. approximately 25% of original volume*

d. more information is needed

9. Gravity thickeners, compared to DAFs, are best suited to:

*a. Thickening primary sludge.

b. Thickening waste activated sludge.

c. Controlling sulﬁde odors.

d. Removing ﬁlamentous bacteria.

e. Provide highest concentration sludge.

10. Which of the following is not the main reason for thickening sludge

a. Improved digester performance due to a lower volume of sludge

b. Cost savings in the construction of new digestion facilities

c. Reduction in anaerobic digestion heating requirements since less water has to be heated

*d. Reduce costs of biosolids hauling

11. What zone is not involved in a belt ﬁlter press?

a. Gravity

b. Low pressure (wedge)

c. High pressure

*d. Twilight

12. The ﬂoat blanket in a DAF unit appears well ﬂocculated and concentrated. Too low a ﬂight speed

would likely result in:

a. Using excessive amounts of air.

b. Float solids that are too thick.

c. Too low an air-to-solids ratio.

*d. Poor thickener underﬂow quality.

e. De-ﬂocculation of ﬂoat solids.

13. The least critical operational control of a dissolved air ﬂotation thickener in producing an ade-

quately thickened sludge is the:

a. Flight speed.

b. Air to solids ratio.

c. Polymer dosage.

*d. Recycle ratio.

e. Pre-thickened WAS concentration.

14. The operational control of a dissolved air ﬂotation thickener most critical for producing an ade-

quately thickened ﬂoat solids is the:

a. Fight speed.

*b. Air to solids ratio.

c. Polymer dosage.

15.4 Dissolved air ﬂoatation thickener (DAFT)

233

d. Recycle ratio.

e. Concentration of WAS being thickened.

15. To increase solids recovery on a dual belt, belt press, the operator should:

a. Increase the differential belt speed.

b. Increase belt speed.

c. Increase sludge feed rate.

*d. Decrease sludge feed rate.

e. Increase polymer feed.

16. Too high a ﬂight speed in a DAF will likely result in:

a. Using excessive amounts of air.

b. Excessive underﬂow volume.

*c. Thin ﬂoat solids.

d. Too high an air to solids ratio.

e. Too low an air to solids ratio.

17. Which one of the following statements is TRUE in regard to DAF thickeners?

*a. The air to solids ratio in a DAF unit is typically 0.03 to 0.05.

b. The speed of the so-called "ﬂights" has very little effect on the concentration of the ﬂoat solids.

c. Adjustments in the air to solids ratio in a DAF unit will affect the ﬂoat solids but not the unit’s

efﬂuent suspended solids.

d. Anionic polymers are typically used to condition the WAS feed to a DAF unit.

18. Which one of the following process units is usually classiﬁed as a sludge thickening device as

opposed to a dewatering device:

*a. DAF unit.

b. Sludge drying bed.

c. Vacuum ﬁlter press.

d. Belt press.

e. All of the above are thickeners, not dewatering devices.

19. Which one the following statements is TRUE in regard to gravity thickeners?

a. Longer solids detention times are desired during summer operation of these units.

b. The sludge-volume-ratio (SVR) or sludge detention time is deﬁned as the volume of the sludge

blanket divided by the daily volume of sludge withdrawn from the thickener.

c. SVRs should be in range of 5 to 10 hours.

*d. A likely cause of a gravity thickener producing a poor quality efﬂuent is too low of a sludge

blanket.*

20. A DAF thickener has efﬂuent solids of 55 mg/L and ﬂoat solids of 2.0%. Solids loading and poly-

mer dosing is in the normal range. This data likely indicates:

a. This unit is operating normally

b. Too low air to solids ratio

c. Float blanket too thick

*d. Flight speed too fast

e. Flight speed too slow

21. An increase in the pool depth of a scroll-type centrifuge:

a. would not affect the moisture content of the cake.

b. would produce a drier cake

*c. would produce a wetter cake, but produce a greater solids recovery.

234

Chapter 15. Sludge Thickening

d. would not affect either solids recovery, nor cake moisture content.

e. would require an increase in the cationic polymer dosage.

22. A sludge thickened from 1% to 4% solids will be reduced in volume by how much?

a. no more than 4% of original volume

b. approximately 17% of original volume

*c. approximately 25% of original volume*

d. more information is needed

23. Gravity thickeners, compared to DAFTs, are best suited to:

*a. Thickening primary sludge

b. Thickening waste activated sludge

c. Controlling sulﬁde odors

d. Removing ﬁlamentous bacteria

e. Provide highest concentration sludge

24. Identify the incorrect statement regarding gravity thickeners.

a. Gravity thickeners are similar in design to primary clariﬁers.

*b. The sludge blanket depth and the rate of sludge withdrawal are used to calculate the sludge

detention time.

c. The purpose of pickets in a sludge thickener is to gently stir settling sludge particles to release

gases that may prevent the sludge particles from compacting.

d. Mixtures of primary and waste activated sludge are never thickened in a gravity thickener due to

the possibilility of de-nitriﬁcation.

e. Solids loading (pounds per day per square foot) is an important guideline for a gravity thickener.

25. On a routine check of a DAFT unit the operator ﬁnds suspended solids of 450 mg/L in the efﬂuent.

The ﬂoat blanket appears well ﬂocculated and concentrated. The operator should:

a. increase the ﬂight speed

b. do nothing, the unit is operating normally

c. reduce the air to solids ratio

*d. increase the air to solids ratio

e. check unit operating pressure

26. Gravity-thickened primary sludge will contain solids within which of the following concentration

ranges?

a. 1,000 - 8,000 mg/l

b. 10,000 – 40,000 mg/l

c. 40,000 - 80,000 mg/l

d. 100,000 - 400,000 mg/l

*e. none of the above

16. Solids Stabilization - Digestion

• Sludge digestion is a microbiological process to treat - stabilize, the wastewater solids

•

As part of digestion, the organic material present is biodegraded by microorganisms. It is the most

common sludge stabilization method.

•

There are two major sludge digestion processes - aerobic digestion which utilizes aerobic microor-

ganisms and anaerobic digestion which utilizes anaerobic microorganisms.

Comparison of aerobic and anaerobic digestion:

Methane production (fuel use potential)

Energy consumption

Unlike anaerobic digestion, aerobic digestion does not produce methane

Aerobic digestion has higher energy needs as air supply is required to

support the aerobic activity. The energy needs for an anaerobic digester is

limited to mixing the digester and maintaining digester temperatures

Anaerobic digestion produces only about 20% of the biomass produced by

Biomass production

aerobic digestion

Dewaterability of sludge produced

Compared to anaerobic digestion, aerobic digestion produces sludge

with poor dewaterability which leads to producing cake with lower solids

content and therefore more expensive to haul

Siginifcant ammonia is produced as part of anaerobic digestion which

Ammonia concentration

imposes additional ammonia removal needs particularly for plants with

nitrogen discharge thresholds

Sludge odors

Capital cost

Aerobic digestion has much less odor issues compared to anerobic diges-

tion

Initial capital costs are higher for anaerobic digesters due to higher SRT

requirements is higher - methanogens are slower growing

Process efﬁciency of aerobic digester drops in cold weather. The anerobic

Weather impacts

digestion is not affected by weather as the sludge in the digester is main-

tained at the desired temperature

For this class we will focus only on anaerboic digestion - the most common method - as anaerboic diges-

tion generates digester gas which can be used as fuel and also due to its lower energy demands.

236

Chapter 16. Solids Stabilization - Digestion

Figure 16.1: Floating Dome Anaerobic Digestion

16.1 Digester design

• The anaerobic digester is typically a large cylindrical concrete tank

• The digester is operated as a continuous process at a ﬁxed volume

•

As sludge is fed into the digester it displaces an equal amount of sludge which leaves the digester

through the digester overﬂow system - see the digester overﬂow system design diagram below

The digester overﬂow system is designed to allow for selecting the level from which the sludge

overﬂows

The sludge typically occupies 70 - 90% of the total digester volume and the methane carbon diox-

ide gas mixture occupies the headspace from where it is withdrawn also on a continuous basis.

• The digester can be constructed with either a ﬁxed or ﬂoating roof.

16.2 Types of anaerobic digestion systems

Low-Rate

• No mixing provided

• Long detention time of 30 to 60 days

• Suitable for low organic loading

• Intermittent sludge feed

•

Stabilized solids settle to the bottom and are re-

moved periodically

• Suitable for small wastewater treatment plants

16.2 Types of anaerobic digestion systems

237

Figure 16.2: Fixed Dome Anaerobic Digestion

Figure 16.3: Digestion Overﬂow System

238

Chapter 16. Solids Stabilization - Digestion

High-Rate

• Mixing and external heating provided

• Uniform feeding

• Typical detention times of 10 to 20 days

• Most common

Two-stage

•

Digestion occurs in the heated and

mixed primary digester vessel

•

The unheated secondary tank

serves as storage for digested solids

and as a source of seed

Summary Conditions for Anaerobic Sludge Digestion

Temperature and detention

Psychrophillic

50 - 65 F 50 - 180 days

time

Mesophillic - Normal range

Mesophilic Most common

68 - 113 F

95 - 98 F 15 - 30 days

Thermophillic

Optimal

113 - 135 F 5 - 12 days

7.0 to 7.1

General Range

Optimal

6.4 to 7.4

VS loading rate (high rate

0.1 and 0.2 lbs VS/day/ ft³

digester)

Gas production

Per pound volatile solids added

8-12 cu. ft

Per pound volatile solids destroyed

Methane

16-18 cu. ft.

65%

Gas Composition

Carbon dioxide

35%

Hydrogen sulﬁde

Normal

Trace

200-800 mg/L

Volatile acid concentration

(as acetic acid)

Alkalinity Concentration

Maximum

Normal

2000 mg/L

2,000-3,500 mg/L (as calcium carbonate)

3,000-5,000 mg/L (as bicarbonate)

Volatile acids to alkalinity

ratio

Desirable

< 0.25

Upset condition

> 0.4

16.3 Digestion parameters and control

239

16.3 Digestion parameters and control

16.3.1 Digestion temperature

•

The anaerobic digestion is temperature sensitive and temperature of the sludge in the digester needs

to be controlled to ensure proper operation.

The activity and type of bacteria present in the digester is dictated by the operating temperature of

the digester.

Anaerobic digestion can be in the following three temperature ranges, each of which has its own

unique microbiology.

i. PSYCHROPHILIC DIGESTION

If operating in this temperature regime, the digesters are maintained between 50 - 65 F. This

regimen is not commonly used. Digestion in this range requires 50 to 180 days depending on

the target volatile matter reduction

ii. MESOPHILIC DIGESTION:

These digesters operate between 68 - 113 F. Most common operating temperatures for

mesophilic digesters is between 95 – 98 F and the typical number of days required for di-

gestion is between 15 to 30 days.

iii. THERMOHILLIC DIGESTION:

These digesters’ optimal operating temperatures range is between 113 - 135 F and it typically

requires 5 to 12 days.

•

A digester does not tolerate temperature ﬂuctuations and need to be maintained at stable tempera-

tures for a signiﬁcant duration.

• As a rule of thumb temperature of a digester should not be changed more than 1 per day.

• Temperature ﬂuctuations will destroy the microbiological balance leading to digester failure.

• External heating is provided to maintain sludge temperatures in the digester.

• Temperature ﬂuctuations also can be minimized by feeding the system at frequent intervals.

•

An external or internal heat exchanger or direct steam injection is typically used for maintaining the

sludge temperature

16.3.2 Digestion mixing

A sludge mixing system is integral to the digester design. It performs three main functions:

1. It allows for the bacteria to properly come in contact with the food.

2. It prevents accumulation of grit and scum in the digester which could lead to dead zones.

3. It allows for maintaining uniform temperature. Typical mixing system designs achieve a turnover

time of 30 to 45 minutes.

16.3.3 Digestion feed

• The sludge feed to the digester has a total solids content of about 3 – 6%.

• 70% of the total solids fed are organic solids.

• These organic solids are measured as volatile solids (VS)

•

Hydraulic or Sludge retention time (SRT) is calculated by dividing the digester volume by the daily

sludge ﬂow

Digester Volume (ft³or gallons)

Digester SRT (days) =

Sludge Feed ( ft³or gallons per day)

240

Chapter 16. Solids Stabilization - Digestion

• The SRT in an anaerobic digester typically ranges from 10 to 20 days.

•

A minimum SRT is essential to the digestion process to ensure that the necessary microorganisms

are being produced at the same rate as they are removed from the system each day

Accumulation of grit in the bottom or scum at the top of the digester would effectively reduce the

working volume of the digester

The amount of digester inﬂuent being pumped and the percent VS of the waste are a measure of the

digester’s organic loading rate (inﬂuent mass per time)

Control of the amount of volatile solids loading is one of the key critical operational control param-

eter.

Volatile solids loading rate of the digester is the mass of VS added to the digester each day divided

by the operating volume of the digester – lbs VS/day/ft³

• Typical high rate digesters VS loading range between 0.1 and 0.2 lbs VS/day/ ft³

• The rate of digestion depends on the type of organic material present.

•

The constituents of the organic material in primary sludge is typically feces, and plant and animal

origin material. This organic material is relatively easy to digest compared to the organics in the

secondary sludge which is primarily living and dead microorganisms and metabolic byproducts

associated with microbiological growth and decay.

Typical primary sludge from a treatment plant with preliminary treatment contains approximately

65% organic material. 80% of this organic material is typically destroyed in the digester. Whereas,

the secondary sludge typically contains 90% organics and only 60% of the secondary sludge organ-

ics is destroyed.

16.3.4 Volatile solids breakdown

•

The volatile solids content of the sludge entering and leaving the digester are measured to quantify

the solids removal in the digester

The expected reduction of volatile solids of a properly operating digester is 40-60% of the total

volatile solids present in raw sludge feed.

The volatile solids in the feed sludge would be about 70% to 75% while the digested sludge would

be 45% to 50% volatile solids

The expected reduction of volatile solids of a properly operating digester is 40-60% of the total

volatile solids present in raw sludge feed

• The volatile solids reduction of the digester is provided by the Van Kleeck equation

VS_in−VS_out

Digester VS reduction(%) =

∗100

VS_in−VS_in∗VS_out

•

Digester volatile solids concentration is typically expressed as a percentage of the sludge total

solids

• 70% VS which means that 70% of the total solids is volatile solids

•

The value of VS_inand VS_outfor the digester VS reduction (Van Kleek) equation above should be in

fraction and not as a percentage.

•

For example: for a 70% VS content use 0.7 in the equation. Likewise 0.525 is used for 52.5% VS

concentration

• Higher volatile solids reduction implies higher gas production and lower biosolids hauling costs

•

Breakdown of volatile matter in the sludge ultimately into methane (CH₄) and carbon dioxide

(CO₂) occurs in multiple steps involving different groups of microorganisms.

16.3 Digestion parameters and control

241

Step 1. This is the hydrolysis step where the complex organic matter in the sludge including carbo-

hydrates, proteins, lignin, and lipids are converted to simpler compounds including sugars, soluble

fatty acids and amines.

Step 2. This involves formation of volatile acids from the products from Step 1. This is the acid

formation step.

Step 3. The acid formed in Step 2 is converted into methane and carbon dioxide. This step is ac-

complished by a group of methane forming bacteria.

•

The digester conditions are maintained to ensure optimal conditions conducive to the activity of the

methane formers.Good digester operations requires ensuring that conditions are kept favorable for

the methane formers

16.3.5 Digester pH and alkalinity

•

The anaerobic digestion requires a symbiotic relationship between the different groups of microor-

ganisms involved.

Each group of organisms involved in the anaerobic digestion process have an optimal pH for maxi-

mum rate of reaction –

Hydrolysis: pH 5-7 optimal

Acid formation: pH 5-7 optimal –

Methane formation: pH 7-8 optimal, pH 6.5-8.5 operational

•

In a healthy digester the volatile acids produced in Step 2 are used in Step 3 as food by the methane

formers at about the same rate as they are produced.

As the methane forming microorganisms in the digesters are very sensitive to the digester con-

ditions including organic content, pH, temperature and toxins and also grow very slowly, it is

necessary to maintain the digester pH close to neutral

•

The conversion of the volatile fatty acids to methane by methane formers allow for controlling the

accumulation of volatile acids

If the activity of methane formers is suppressed due to conditions such as low pH, volatile acids

will accumulate potentially causing the digester to become "sour"

In a normally operating digester, the alkalinity present in the sludge prevents the pH from dropping

due to the formation of volatile acids during Step 2.

The alkalinity consumed due to the acids formed in Step 2 is replenished in Step 3 by the formation

of bicarbonate from the CO₂and ammonia produced during digestion.

NH₃+H₂O+CO₂→ NH₄HCO₃

•

Unlike the acid formers, Thus, if the digester condition such as the pH was to change – drop be-

cause of the formation of acids and a lack of alkalinity, Step 3 will not happen resulting in the

digester pH dropping even further resulting in a “sour” digester condition.

• When acid formers out produce the methane formers the volatile acids increase sharply

•

The methane formers are the key to digester operation and are very vulnerable to low pH There

needs to be sufﬁcient alkalinity present to prevent pH changes as the volatile fatty acids are being

formed.

•

If sufﬁcient alkalinity is not present the pH will drop further inhibiting the activity of methane

formers and the digester will turn “sour”

The bicarbonate ion (HCO⁻₃) is the main source of buffering capacity to maintain the system’s pH

242

Chapter 16. Solids Stabilization - Digestion

in the range of 6.5 – 7.6. The concentration of HCO⁻₃in solution is related to the percent of carbon

dioxide in the gas phase

•

Alkalinity consumed by the acid production is offset by alkalinity produced as part of the methanogenic

activity

Regular monitoring the digester alkalinity and volatile acids is vital in ensuring a stable digestion

process.

In a well operating digester the volatile acids expressed as acetic acid would be in the range of 50

to 500 mg/L and the alkalinity expressed as calcium carbonate would be in the range of 2,000 to

3,000 mg/L (bicarbonate alkalinity of 2,500mg/L and 5,000mg/L)

The volatile acid-to-alkalinity ratio indicates if the digester has enough buffering capacity for the

volatile acids being produced.

•

Increase in volatile acid to-alkalinity ratio will increase the potential for pH decrease which would

result in an upset digester.

In general keeping the volatile acids to alkalinity ratio at 0.25 or less is desirable for good opera-

tions.

• Ratios above 0.4 indicate upset and the need for corrective action.

16.3.6 Digester pH control

•

Following chemicals: calcium oxide, calcium hydroxide, anhydrous ammonia, Ammonia (liquid),

ammonium hydroxide, sodium carbonate, sodium bicarbonate, and sodium hydroxide have been

used for increasing the pH of a sour digester.

The use of calcium oxide (quicklime) or calcium hydroxide (slaked or hydrated lime) is probably

the most dangerous if mixed with water prior to feeding. If instead of adding these two chemicals

to water for digester feeding, adding water to these chemicals would pose the threat of a violent

reaction and splattering of this strong caustic.

Excess lime should be avoided as lime reacts with carbon dioxide to form calcium carbonate.

Ca(OH)₂+CO₂→ CaCO₃+H₂O

If carbon dioxide is removed too rapidly or in too large a quantity from the sludge, then carbon

dioxide from the biogas will replace the carbon dioxide lost from the sludge. When carbon dioxide

is lost from the biogas, a partial vacuum condition develops under the digester dome. This con-

dition may cause the digester cover to collapse. Also, as the concentration of alkalinity increases

in the anaerobic digester, the continued use of quick lime results in the precipitation of calcium

carbonate.

• Sodium hydroxide would be the next most dangerous followed by ammonium hydroxide.

•

The use of anhydrous ammonia should be limited to facilities that have the equipment to handle gas

cylinders and that have feed connections or the ability to make the necessary feed connections.

The feeding of lime can only raise the digester pH to about 6.8. When lime is added it reacts with

carbon dioxide to form calcium bicarbonate. Excessive feeding of lime causes insoluble calcium

carbonate to form. In addition excessive lime may remove too much carbon dioxide which could

lower gas pressures or even the formation of a vacuum. This could cause air to be drawn into the

digester and cause an explosive gas mixture.

•

16.3.7 Digester gas production

In the anaerobic digestion process microorganisms convert volatile matter into mainly methane

(CH₄) and carbon dioxide (CO₂)

•

16.4 Digester safety

243

•

Typical composition of digester gas is about 58 - 65% methane and 30 - 35% CO2 it also includes

traces of ammonia nitrogen, hydrogen sulﬁde, and other gases.

• Monitoring digester gas production is one of the more important operational parameter.

•

Gas production ranges between 10 to 16 cubic feet per pound of volatile matter destroyed and the

gas production remains stable over time.

•

Low gas production is an indicator of digester issue related to - toxicity, temperature, volatile acid

to alkalinity ratio, mixing, or feed rates.

16.3.8 Digester failure

• Digester failure conditions are marked by conditions which include increases in:

– volatile acids

– volatile acids: alkalinity ratio

– drop in pH

– drop in gas production, and

– an increase in CO₂concentration

•

The failure is typically attributable to factors including overloading, toxicity and temperature

control issues.

• Digester failure can be controlled by:

– stopping or reducing the sludge feed

– exercising pH controls by adding alkaline chemicals, and

– through reseeding the digester with sludge from another healthy digester

16.3.9 Digester toxicity issues

• Toxicity could severely impact a digester performance.

•

Contributors to toxicity issues include high concentrations of ammonia and toxic metals such as

arsenic, cadmium, chromium (hexavalent), copper, nickel, zinc.

•

Ammonia toxicity: Ammonia from industrial sources, organic overloading, or from using anhy-

drous ammonia to correct a digester pH problem can cause a die-off of bacteria; ammonia toxicity

begins around 1,500 mg/L and is totally toxic at 3,000 mg/L.

16.3.10 Nutrients and trace constituents requirement

The major nutrients required for anaerobic digestion are nitrogen and phosphorus. An average cell con-

tains approximately 12.5 percent nitrogen and 2 percent phosphorus. Sodium, potassium, calcium, magne-

sium, chloride, and sulfate ions are also required for proper digestion.

16.4 Digester safety

A. Pressure & Vacuum Protection:

•

In a digester,as the gas production and sludge withdrawal is continuous, there exists a poten-

tial to create pressurized or vacuum conditions within the digester which could jeopardize the

structural integrity of the digester.

•

Typically the gas in the digester dome is maintained at a pressure of 8-9 inches of water

column.

A pressure relief valve is designed into the digester dome which allows to protect the digester

from excessive pressure or vacuum conditions.

244

Chapter 16. Solids Stabilization - Digestion

•

In the event of the digester dome being under excessive pressure, the valve will open to

relieve the pressure and likewise if the valve senses a low pressure or vacuum condition, it

will open up just enough to let in the air to bring the gas in the digester dome back to normal,

set pressure.

B. Fire and Explosion Protection:

• For a ﬁre or explosion to occur, following three conditions must be met simultaneously:

(a) Presence of fuel – combustible material

(b) Oxygen (air) must exist in certain proportions, and

• Digester gas contains methane which is a combustible gas.

•

The ratio of fuel and oxygen that is required varies with each combustible gas or vapor. The

Lower Explosive Limit (LEL) and (UEL) for methane is 5% and 15% respectively.

A 5% LEL implies that a methane concentration in air of less than 5% is too “lean” to burn

and a 15% UEL implies that a concentration of methane exceeding 15% in air would be too

“rich” to burn. So for methane, the ﬂammable range is between 5% and 15%. As the digester

gas methane content is typically between 60 to 65%, dilution of digester gas with air has the

potential to form an explosive mixture.

•

A ﬂame arrestor is typically installed at the outlet of the digester gas piping to protect the

digester from any external heat or ignition source. The ﬂame arrestor is essentially a heat

exchanger which cools down the ignition source or ﬂame entering the digester.

16.4 Digester safety

245

Digester Pressure Relief

Digester Flame Arrestor

Chapter Assessment

1. A change in the-pH-of digesting sludge in anaerobic digester is the best early warning indicator of

potential digester upset.

a. True

*b. False

2. A good maintenance program should be established for all ﬂame arresters to ensure they are all set

at the recommended "pop-off’ pressures.

*a. True

b. False

3. A healthy anaerobic digester should have a carbon dioxide concentration of more than 40%.

a. True

*b. False

4. A high rate anaerobic digester is always heated and mixed.

*a. True b.False

5. Anaerobically digested sludge produces gas as a by-product. The gas produced is of little or no

value.

a. True

*b. False

6. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or

ﬂame.

a. True

*b. False

7. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or

ﬂame

*a. True

b. False

8. In an anaerobic digester, the [1] is destroyed producing [2]. The digester gas production is typically

248

Chapter 16. Solids Stabilization - Digestion

in the range of [3] cubic feet of digester gas per lb of volatile matter destroyed.

9. Gas production in an anaerobic digester results from the destruction of ﬁxed solids

a. True

*b. False

10. A well operating digester will have a CO2 concentration of greater than 60%

a. True

*b. False

11. Volatile solids removal efﬁciency of a digester is most commonly monitored utilizing the BOD test

a. True

*b. False

12. Volatile solids removal efﬁciency of a digester is most commonly monitored utilizing the BOD test

a. True

*b. False

13. The range of volatile material in raw sludge may be as high as 60% - 80% while digested sludge

normally will be below 20%.

a. True

*b. False

14. Foaming in digesters is often due to rapid acid digestion is caused by adding too much raw sludge

during too short a time period.

*a. True

b. False

15. Aerobic digestion produces methane gas that provides energy for other operations.

a. True

*b. False

16. A good maintenance program should be established for all ﬂame arresters to ensure they are all set

at the recommended "pop-off" pressures.

a. True

*b. False

17. A ﬂame arrester in the gas line between a waste gas burner and an anaerobic digester contains

plates which should periodically be inspected and cleaned.

*a. True

b. False

18. The methane forming bacteria in an anaerobic digester reproduce more rapidly than the acid form-

ing bacteria.

a. True

*b. False

19. The main function of a secondary digester in normal operation is to provide a storage space for

seed sludge.

a. True

*b. False

20. Supernatant liquor is withdrawn from the lowest possible level in an anaerobic digester.

a. True

*b. False

21. The pH of digested sludge in a healthy anaerobic digester will be near 7.0

*a. True

16.4 Digester safety

249

b. False

22. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.

*a. True

b. False

23. A properly operated anaerobic, digester functions best within a pH range of 5.4 to 5.8.

a. True

*b. False

24. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the

supernatant system.

a. True

*b. False

25. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will

occur.

*a. True

b. False

26. Aerobic digesters are operated under the principle of extended aeration from the activated sludge

process.

*a. True

b. False

27. After pumping raw sludge to a digester, it is necessary to seal both ends of the line to prevent back

siphonage.

a. True

*b. False

28. Propeller mixers are located mainly on ﬂoating covers of anaerobic digesters rather than on ﬁxed

covers.

a. True

*b. False

29. A gallon of water weighs 8.34 pounds. A gallon of digested sludge, due to its high volatile content,

will weigh much less.

a. True

*b. False

30. The range of volatile material in raw sludge may be as high as 60% - 80% while digested sludge

normally will be below 20%.

a. True

*b. False

31. Foaming in digesters is often due to rapid acid digestion is caused by adding too much raw sludge

during too short a time period.

*a. True

b. False

32. Aerobic digestion produces methane gas that provides energy for other operations.

a. True

*b. False

33. A good maintenance program should be established for all ﬂame arresters to ensure they are all set

at the recommended "pop-off" pressures.

a. True

250

Chapter 16. Solids Stabilization - Digestion

*b. False

34. A ﬂame arrester in the gas line between a waste gas burner and an anaerobic digester contains

plates which should periodically be inspected and cleaned.

*a. True

b. False

35. The methane forming bacteria in an anaerobic digester reproduce more rapidly than the acid form-

ing bacteria.

a. True

*b. False

36. The main function of a secondary digester in normal operation is to provide a storage space for

seed sludge.

a. True

*b. False

37. Supernatant liquor is withdrawn from the lowest possible level in an anaerobic digester.

a. True

*b. False

38. The pH of digested sludge in a healthy anaerobic digester will be near 7.0

*a. True

b. False

39. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.

*a. True

b. False

40. A properly operated anaerobic, digester functions best within a pH range of 5.4 to 5.8.

a. True

*b. False

41. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the

supernatant system.

a. True

*b. False

42. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will

occur.

*a. True

b. False

43. Aerobic digesters are operated under the principle of extended aeration from the activated sludge

process.

*a. True

b. False

44. After pumping raw sludge to a digester, it is necessary to seal both ends of the line to prevent back

siphonage.

a. True

*b. False

45. A gallon of water weighs 8.34 pounds. A gallon of digested sludge, due to its high volatile content,

will weigh much less.

a. True

*b. False

16.4 Digester safety

251

46. Gas production in an anaerobic digester results from the destruction of ﬁxed solids in the raw

sludge.

a. True

*b. False

47. "High rate" and Low rate" when used in referring to an anaerobic digester may refer to the rate of

volatile solids loading.

*a. True

b. False

48. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the

supernatant system.

a. True

*b. False

49. Increased concentrations of volatile acids and decreased alkalinity are the ﬁrst measurable changes

that take place when the process of anaerobic digestion is becoming upset.

*a. True

b. False

50. A change in the-pH-of digesting sludge in anaerobic digester is the best early warning indicator of

potential digester upset.

a. True

*b. False

51. A good maintenance program should be established for all ﬂame arrestors to ensure they are all set

at the recommended "pop-off’ pressures.

*a. True

b. False

52. A healthy anaerobic digester should have a carbon dioxide concentration of more than 40%.

a. True

*b. False

53. A high rate anaerobic digester is always heated and mixed.

*a. True

b. False

54. Anaerobically digested sludge produces gas as a by-product. The gas produced is of little or no

value.

a. True

*b. False

55. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or

ﬂame.

a. True

*b. False

56. Volatile solids removal efﬁciency of a digester is most commonly monitored utilizing the BOD test

a. True

*b. False

57. Gas production in an anaerobic digester results from the destruction of ﬁxed solids in the raw

sludge.

a. True

*b. False

252

Chapter 16. Solids Stabilization - Digestion

58. "High rate" and "Low rate" when used in referring to an anaerobic digester may refer to the rate of

volatile solids loading.

*a. True

b. False

59. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the

supernatant system.

a. True

*b. False

60. Increased concentrations of volatile acids and decreased alkalinity are the ﬁrst measurable changes

that take place when the process of anaerobic digestion is becoming upset.

*a. True

b. False

61. In a two-stage anaerobic digestion system, it is in the secondary tank where most of the volatile

solids destruction occurs

a. True

*b. False

62. The lower explosive limit for methane is 40%

a. True

*b. False

63. Volatile solids removal efﬁciency of a digester is most commonly monitored utilizing the BOD test

a. True

*b. False

64. Sludge digestion is a process by which principally the inorganic matter in sludge is gasiﬁed and/or

converted to a more stable form by the biological process.

a. True

*b. False

65. The determination of the pH on the anaerobic digester is one of the best early warning systems of

digester upset.

a. True

*b. False

66. The gas produced as part of the anaerobic digestion is of little or no value

a. True

*b. False

67. The main function of a secondary digester in normal operation is to provide a storage space for

seed sludge.

a. True

*b. False

68. The pH of digested sludge in a healthy anaerobic digester will be near 7.0.

*a. True

b. False

69. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.

*a. True

b. False

70. Volatile solids removal efﬁciency of a digester is most commonly monitored utilizing the BOD test

a. True

16.4 Digester safety

253

*b. False

71. Identify the incorrect statement regarding anaerobic digestion.

a. An anaerobic digester with pH of 7.05, an alkalinity of 2,900 mg/l and a volatile acid concentra-

tion of 250 mg/l is probably operating normally.

*b. Sodium bicarbonate may be used in place of lime to neutralize a sour anaerobic digester.

c. When adding lime to a sour anaerobic digester, it is important to add an excess of this chemical

to act as a reservoir of alkalinity.

d. Ferrous sulfate may be added to an anaerobic digester to reduce hydrogen sulﬁde concentration

where air quality is of concern.

e. Gas production from anaerobic digester may be expressed as cubic feet of gas produced per

pound of volatile matter added per day.

72. One action that may be taken to improve the health of an anaerobic digester that is going sour

would be to:

a. Add small doses of lime daily to maintain the digester pH above 7.0.

*b. Add ferrous sulfate to reduce the concentration of hydrogen sulﬁde in the digester.

c. Add seed sludge from a healthy primary digester.

d. Pump raw sludge to this digester more frequently so that this sludge has a higher pH.

e. Increase the temperature of the digester to favor a population increase of the methane formers.

73. Identify the incorrect statement regarding operation of an anaerobic digester.

*a. A healthy anaerobic digester would generally have volatile acids in the range of 50 mg/l to 300

mg/l.

b. The best strategy for pumping raw sludge to an anaerobic digester is to pump it once or twice a

day so that the thickest possible raw sludge may be pumped.

c. An anaerobic digester operating at an alkalinity of 3200 mg/l should be able to tolerate a volatile

acid concentration of 250 mg/l.

d. The change in pH is not a reliable indicator of the changing characteristics of the digesting

sludge because the alkalinity in an anaerobic digester acts as a buffer.

e. The higher the volatile solid content of the primary sludge being fed to the digester, the higher is

the expected volatile reduction.

74. Identify the true statement about anaerobic digesters:

a. Carbon dioxide and methane in digester gas should be 65% and 35%, respectively.

*b. Increasing carbon dioxide readings indicate possible organic overload.

c. Decreasing mixing will always improve recovery of a sour digester.

d. Increasing the frequency and decreasing the amount of sludge pumping will not improve digester

performance.

e. Reducing the ratio of primary to waste activated sludge will improve gas production.

75. All of the following are normal operating guidelines for a healthy anaerobic digester except for:

a. A mesophilic digester operating at 93°F to 98°F.

b. Methane gas in the range of approximately 62% to 70%.

c. Carbon dioxide gas in the range of 30% to 38%.

*d. Organic loading to a high rate digester of 0.15 to 0.2 pounds (Lb) volatile Solids per day per ft3

of digester capacity.

e. Bicarbonate alkalinity in the range of 1500 to 1800.

76. When adding anhydrous ammonia to a sour primary anaerobic digester, the

tant consideration because if it is too high, digester "poisoning" may result:

is an impor-

254

Chapter 16. Solids Stabilization - Digestion

a. Concentration of hydrogen sulﬁde in digester gas.

*b. pH in digesting sludge.

c. Concentration of free copper ions in digesting sludge

d. Ferrous and ferric ion concentration in digesting sludge.

e. Dissolved sulﬁde concentration.

77. Which one of the following parameter ranges is most appropriate for a healthy anaerobic digester?

a. Volatile solid reduction in the range of 60 to 70%.

b. Digester gas production of 7 to 10 ft3 of gas produced per day per pound of volatile solids de-

stroyed.

*c. Alkalinity in the range of 2500 to 3500 mg/L.

d. Hydrogen sulﬁde in range of 1% to 2% by volume.

e. Volatile acids in range of 500 to 750 mg/L.

78. Identify the incorrect statement regarding operation of an anaerobic digester.

a. A healthy anaerobic digester would generally have volatile acids in the range of 50 mg/l to 300

mg/l.

*b. The best strategy for pumping raw sludge to an anaerobic digester is to pump it once or twice a

day so that the thickest possible raw sludge may be pumped.

c. An anaerobic digester operating at an alkalinity of 3200 mg/l should be able to tolerate a volatile

acid concentration of 250 mg/l.

d. The change in pH is not a reliable indicator of the changing characteristics of the digesting

sludge because the alkalinity in an anaerobic digester acts as a buffer.

e. The higher the volatile solid content of the primary sludge being fed to the digester, the higher is

the expected volatile reduction.

79. Lab data from your 100,000 gallon primary anaerobic digester, which receives primary sludge only,

is shown below. Using this data :

Date

Alkalinity(mg/L)

3,200

Vol.Acids(mg/L)

CO₂(%)

35.5

9/02

7.10

280

9/09

9/16

9/17

7.00

6.90

6.85

3,020

2,800

2,720

320

400

450

36.0

37.7

38.2

Date

RawSludge(TS%)

5.4

Raw Sludge(VS%)

65.5

Digested Sludge(VS%)

56.0

9/02

9/09

9/16

5.0

4.9

66.7

65.9

53.8

54.2

(a) Calculate the average volatile solids reduction. Compare your calculated value to generally

accepted ranges for a healthy anaerobic digester. Comment.

(b) Compare the other data to expected ranges.

16.4 Digester safety

steps that may be taken to mitigate the problem.

255

(d) Should slake lime be added ? Why or why not ?

80. As part of an ongoing training program at your wastewater plant you are assigned the task of

preparing a lecture for OITs on the subject of "buffer" in an anaerobic digester. Answer the fol-

lowing questions in preparation for your presentation.

a. What is "buffer" and why is it important in an anaerobic digester?

b. What lab test is used to measure "buffer" in an anaerobic digester? Explain how this test is

run

c. Most of the buffer in an anaerobic digester is due to the presence of

17. Solids Dewatering

•

Dewatering, like thickening does not treat the sludge but it allows for a reduction in sludge volume

by removing water.

Thickening achieves about 10% or less solids content while dewatering is typically for increasing

the solids content to between 15 to 30 percent.

Sludge is dewatered to make it easier to handle and to reduce costs associated with elements re-

lated to accomplishing the end objectives with the sludge – land application, composting, drying,

incineration or landﬁll.

•

Dewatering involves conditioning the sludge with a polymer and subjecting it to a physical process

such as belt ﬁlter press or centrifuge to remove the water.

• Reasons for sludge dewatering are:

– Critical for sludge treatment options such as sludge drying and incineration.

– Reduction in the weight of solids to be hauled – reducing hauling cost.

– Reduction in the volume of sludge that needs to be handled

• More common sludge dewatering methods include:

1. Belt Filter Press

2. Centrifuge

• These are both mechanical methods and involve physically removing free water

•

Both these methods involve conditioning of sludge with a cationic polymer which ﬂocculate the

solids - separating it from the water

17.1 Belt Filter Press

The belt ﬁlter press dewaters sludge by squeezing polymer conditioned sludge between a pair of belt

ﬁlter fabric as these belts pass through a system of rollers. Belt ﬁlter press produces a dewatered product

typically between 12% to 35% solids content.

258

Chapter 17. Solids Dewatering

Belt Filter Press

17.1.1 Principles of belt ﬁlter press operations

•

Belt ﬁlter press utilizes two endless, porous belts (generally 0.7 to 3 meter wide) made from a

synthetic material

Polymer ﬂocculated sludge is ﬁrst introduced on the horizontal gravity zone of the press where free

water is removed from the sludge by chicane assisted gravity ﬁltration through the porous belt press

fabric.

17.1 Belt Filter Press

259

Gravity Zone - Chicanes

•

From the gravity zone, the sludge enters the low pressure zone (wedge zone) in which the sludge is

prepared for the upcoming high-pressure zone by evenly distributing the solids across the belt and

gradually applying pressure

In the high pressure zone the sludge is sandwiched between two belt press fabrics. As the belts

with the sandwiched sludge passes over and under 6 to 12 tensioning rollers which progressively

decrease in diameter. The applied pressure squeezes the water out from the sludge and is removed

through the porous fabric.

• The ﬁltrate is returned back to the front of the plant.

•

The belts are washed continuously as they pass over wash boxes. The wash boxes are equipped with

rotating brushes and water sprays and are primarily for keeping the belt press fabric pores open so

water could easily ﬁlter through.

• Improper polymer dosage and excessive hydraulic loading can lead to a wetter sludge cake.

•

The belts are operated at speeds that commensurate with the rate of solids loading and the type

of sludge being dewatered. Lower belt speed would allow for better water removal. However, if

if the belts speed is below the threshold for that particular sludge, belt press washout, which is

characterized by sludge ﬂowing over the sides of the belt can occur . Thus, the best belt speed is the

slowest speed the belt can be operated without causing washout.

•

The belts can be blinded - loose its porosity causing the water to stagnate/cause washout and not

ﬁlter through, if excessive polymer is used or due ineffective belt washing.

Solids capture rate and Percent cake solids produced are the most important belt press efﬁciency

measurement parameters

17.1.2 Elements of the belt ﬁlter press

• belts

• wash boxes

• guiding and tensioning rollers

• chicanes

• doctor blades

• drive motor

• hydraulic unit

• polymer injection and mixing system

260

Chapter 17. Solids Dewatering

17.1.3 Belt press operational parameters

• belt ﬁlter width

• belt ﬁlter speed

• hydraulic loading

• belt tension

• washout

• ﬁltrate quality

• solids capture rate

• polymer dosing rate (lbs polymer/dry ton solids)

• solids and hydraulic loading rates, and

• quantity and characteristics of the polymer used

17.2 Centrifuge

The centrifuge dewaters the sludge by subjecting a polymer conditioned sludge to strong centrifugal

forces by rotating it in a bowl at high speeds. Centrifuges are used for dewatering solids in many differ-

ent applications and wastewater solids dewatering being one them. Centrifuges also come in different

conﬁgurations. The scroll conveyor (decanter) centrifuge is the more commonly used centrifuge design

for wastewater sludge dewatering. It produces a dewatered product typically between 20% to 30% solids

content. It should be noted that the centrifuge can also be utilized for sludge thickening.

Centrifuge

17.2.1 Principles of centrifuge operations

•

The bowl of the centrifuge is cylindrical shaped with a tapered cone at one end. The bowl is de-

signed to rotate at speeds of 1200 to 2400 RPM.

17.2 Centrifuge

261

Centrifuge Bowl

•

A screw conveyor - scroll is shaped to the bowl contour and carried on a central shaft or drum and

it rotates independently from the bowl.

Centrifuge Scroll

•

The polymer conditioned sludge is fed into the cylindrical portion of the bowl through a central

feed pipe. Inside the bowl, the sludge is subjected to intense G forces around 3000 Gs by virtue of

the bowl rotating at a high speed.

• As the bowl rotates, the water in the ﬂocculated sludge is separated from the solids .

•

The scroll rotates at a 5 to 15 rpm differential speed from the bowl. This differential speed causes

the thickened sludge to be propelled along the bowl for towards its conical end. As the sludge is

screwed up the cone it emerges from the liquid at the beach, is drained and pushed up to discharge

ports.

•

The separated water is discharged from the cylindrical (non-conical) end of the bowl. The depth of

the water - pond depth, is controlled by adjustable weirs at the discharge end.

The greater the pond depth improves the centrate quality and solids recovery but it reduces the

drainage zone at the beach end resulting in wetter solids. The pond depth is altered using adjustable

weirs.

• The centrate is returned back to the front of the plant

17.2.2 Elements of the Centrifuge

• bowl

• scroll

• beach

262

Chapter 17. Solids Dewatering

• adjustable weir

• differential gear box

• drive motor

17.2.3 Centrifuge operational parameters

• bowl speed

• differential speed

• pond height

• centrate quality

• solids capture rate

• polymer dosing rate (lbs polymer/dry ton solids)

• solids and hydraulic loading rates

Chapter Assessment

1. What can be a problem with a belt ﬁlter press?

a. Washing out

b. Polymer overdosing

c. Blinding

*d. All of the above

2. Which one of the following statements is TRUE in regard to the operation of a belt ﬁlter press:

a. Unlike a gravity belt, cationic polymers need not be used to condition the feed sludge to this

unit.

b. When dewatering a mixture of WAS and primary sludge, which has been anaerobically digested,

the operator can expect to produce a sludge cake with 22% TS to 25%.

*c. The best belt speed for this unit is the slowest speed the belt can be operated at, without causing

“washout.”

d. Colloidal solids are likely to cause plugging of belt pores and thus cause “belt binding.”

3. What can be used to evaluate the efﬁciency of a belt ﬁlter press?

a. Vacuum required in inches of mercury

b. % volatile solids in cake

c. Sludge feed rate in gpd

*d. Filter yield in lbs/hr/sq ft

e. Gph of ﬁltrate removal.

4. An increase in the pool depth of a scroll-type centrifuge:

a. would not affect the moisture content of the cake.

b. would produce a drier cake

*c. would produce a wetter cake, but produce a greater solids recovery.

d. would not affect either solids recovery, nor cake moisture content.

e. would require an increase in the cationic polymer dosage.

5. What can be used to evaluate the efﬁciency of a belt ﬁlter press?

264

Chapter 17. Solids Dewatering

a. Vacuum required in inches of mercury

b. % volatile solids in cake

c. Sludge feed rate in gpd

*d. Filter yield in lbs/hr/sq ft

e. gallons-per-hour of ﬁltrate removal.

6. You are assigned to supervise the operation ofa belt ﬁlter press. Anerobically digested sludge - a

mixture of primary and secondary sludge is being dewatered. Lab records show that this belt press

routinely produces a sludge cake having a total solids concentration of 16.5%. No other lab tests

are run on the belt press. Answer the following questions:

a. Are you satisﬁed with the dryness of the cake? Why or why not? Explain your answer.

b. List three (3) variables you could adjust to optimize the BFP. Explain how these adjustments

affect the optimization of this unit.

c. List two (2) other lab tests that must be performed in order to better monitor the operation,of

the BFP

18. Disinfection

18.1 Background

•

The primary goal of water treatment is to ensure that the water is safe to drink and does not contain

any disease-causing microorganisms.

Disinfection refers to an operation to inactivate the microorganisms in water that can cause an

infection or disease. These organisms are collectively referred to as pathogens and include many

species of bacteria, fungus, protozoa, worms, viruses, etc.

•

The processes prior to disinfection - sedimentation and ﬁltration, remove a large percentage of

bacteria and other microorganisms from the water by physical means.

Disinfection is different from sterilization, which is the complete destruction of all organisms

which is expensive and unnecessary.

• Water disinfection can be sub-divided as:

1. Primary disinfection:

–

Kills or inactivates bacteria, viruses, and other potentially harmful organisms in drinking

water.

– Disinfection prevents infectious diseases such as typhoid fever, hepatitis, and cholera

–

Some disinfectants are more effective than others at inactivating certain potentially

harmful organisms.

–

Disinfection processes vary from water utility to water utility based on their needs and

to meet EPA treatment requirements.

2. Secondary disinfection:

–

Maintenance of a disinfectant residual that prevents regrowth of microorganisms in the

water distribution system between treatment and consumer.

Secondary disinfection maintains water quality by killing potentially harmful organisms

such as those that cause Legionnaire’s disease that may get in water as it moves through

pipes.

–

266

Chapter 18. Disinfection

– Monochloramine is commonly used as a secondary disinfectant.

• Elements of an "ideal" disinfectant

– It must act in a reasonable time.

– It must act as temperature or pH changes.

– It must be nontoxic.

– No harmful byproducts.

– It must not add unpleasant taste or odor.

– It must be readily available.

– It must be safe and easy to handle and apply.

– It must be easy to determine the concentration of.

– It must be able to provide residual protection.

– Pathogenic organisms must be more sensitive to the disinfectant than are nonpathogens.

– It must be capable of being applied continually.

– Versatile: effective against all types of pathogens.

– Fast-acting: effective within short contact times

–

Robust: effective in the presence of interfering materials including particulates, suspended

solids and other organic and inorganic constituents

–

Handy: easy to handle, generate, and apply (nontoxic, soluble, non-ﬂammable, non-explosive)

– Compatible with various materials/surfaces in WTPs (pipes, equipments)

– Economical

•

In addition to the desirable characteristics of a disinfectant listed above, the disinfectant chosen

must be able to kill off or deactivate pathogenic microorganisms by one of several possible meth-

ods, including:

1. Damaging the cell wall

2. Altering the ability to pass food and waste through the cell membrane

3. Altering the cell protoplasm

4. Inhibiting the cells’ conversion of food to energy

5. Inhibiting reproduction

18.2 Chlorination

• Despite potential drawbacks, chlorine is the disinfectant of choice.

•

In general, chlorination is effective, relatively inexpensive, and provides effective levels of disinfec-

tant residual for safe distribution.

• Chlorine can be applied as:

– As a gas - elemental chlorine, Cl₂:

– Liquid (sodium hypochlorite)

– Solid (calcium hypochlorite)

each of these forms has advantages and disadvantages.

18.2.1 Chlorine properties

• Chlorine is a yellowish-green gas at room temperature and atmosphric pressure

• Chlorine gas can be pressurized and cooled to its liquid form for making it easy to ship and store.

•

When liquid chlorine is released, it quickly turns into a gas that stays close to the ground (being

heavier than air) and spreads rapidly.

18.2 Chlorination

267

•

While it is not explosive or ﬂammable, as a liquid or gas it can react violently with many sub-

stances

• Chlorine is only slightly soluble in water (0.3 to 0.7% by weight.)

• Chlorine gas has a greenish-yellow color

•

It has a characteristic disagreeable and pungent odor, similar to chlorine-based laundry bleaches,

and is detectable by smell at concentrations as low as 0.2 to 0.4 ppm

• It is about two and a half times as heavy as air

• One volume of liquid chlorine yields about 460 volumes of chlorine gas.

• Liquid chlorine is amber in color and is about one and a half times as heavy as water

• Chlorine is an irritant to the eyes, skin, mucous membranes, and the respiratory system

18.2.2 Chlorine storage and safety

•

Chlorine gas is lethal at concentrations as low as 0.1% air by volume. In nonlethal concentrations,

it irritates the eyes, nasal membranes, and respiratory tract.

Typically for smaller plants chlorine gas is shipped in pressurized steel cylinders - 150 lb or 2000 lb

(ton cylinder) size. Larger plants may get their chlorine supply in rail tank cars.

The daily chlorine usage is typically established based upon the weighing of the chlorine contain-

ers.

The withdrawal rates from a chlorine cylinder is based on the temperature of the liquid in the

cylinder, and thus the pressure of the gas.

• As chlorine gas is withdrawn from the cylinder, it absorbs the heat from the surroundings.

•

For low withdrawal rates, heat will be able to be transferred from the surrounding air to the con-

tainer in time so that there is no drop in temperature or pressure,

If the chlorine withdrawal is larger, the air will not be able to transfer the heat quickly enough and

the temperature (and pressure) of the chlorine will drop, thus resulting in a lower feed rate.

If high enough and prolonged enough, this can even result in ice formation around the outside of

the container, further decreasing the withdrawal rate.

The most effective way to increase withdrawal rate from a single container is to circulate the sur-

rounding air with a fan. Again, never apply heat to the containers.

If chlorine gas escapes from a container or system, being heavier than air, it will seek the lowest

level in the building or area

Only trained staff with access to proper personal protection equipment (PPE) including self-

contained breathing apparatus, should handle the chlorine cylinders and address chlorine leak

issues

•

When a leak is suspected, it is recommended that ammonia vapors be used to ﬁnd the source. When

ammonia vapor using a rag or brush, is directed at a leak, a white cloud will form. To produce am-

monia vapor, a plastic squeeze bottle containing about 5 % ammonia, aqua ammonia (ammonium

hydroxide solution) should be used. A weaker solution such as household ammonia may not be

concentrated enough to detect minor leaks

•

All safety equipment should be located outside of the chlorine room and be easily accessed by all

personnel

Small leaks around valve stems can usually be corrected by tightening the packing nut or closing

the valve. A leak can also be reduced by removing the chlorine as rapidly as possible

If it cannot be added to the process there are several chemicals which can be used to absorb the

chlorine gas. For example, chlorine can be absorbed by using 1frac14 pounds of caustic soda or

268

Chapter 18. Disinfection

hydrated line, or 3 pounds of soda ash per pound of chlorine.

•

If the leaking container can be moved, it should be transported to an outdoors area where minimal

harm will occur. Keep the leaking part the most elevated so that gaseous chlorine will leak rather

than liquid chlorine.

If the leak is large, all persons in the adjacent area must be warned and evacuated. Only authorized

persons equipped with the proper breathing apparatus, and protective measures to the eyes and

body should investigate.

As water is not an efﬁcient absorbent for chlorine and the fact that chlorine reacts with water to

form very corrosive hydrochloric acid, never apply water to a leak or consider submerging a chlo-

rine cylinder (for example, in a pond or tank), since it will probably ﬂoat.

• Remember to keep windward of the leak.

•

As chlorine cylinders pressure increases with temperature, as a safety measure the chlorine cylin-

ders are ﬁtted with fusible plug which melts between 158^oand 165^oF.

Keep chlorine cylinder or container emergency repair kits available. Be familiar with their use and

location.

Leaks at fusible plugs and cylinder valves requires special handling and emergency equipment. The

chlorine supplier must be notiﬁed immediately

Pin hole leaks in cylinder walls or ton tanks can usually be stopped by mechanical pressure applica-

tions (clamps, turnbuckles, etc.). This only temporary and may require your ingenuity.

• Leaking containers cannot be shipped.

• In general, daily inspection of all chlorine cylinders will avoid major problems

18.2.3 Forms of chlorine

Due to safety issues related to the use of chlorine gas, hypochlorites are often used in lieu of

chlorine

•

• Types of hypochlorites

– Sodium hypochlorite (NaOCl) comes in a liquid form which contains up to 12.5% chlorine

–

Calcium hypochlorite (Ca(OCl)₂), also known as High-test Hypochlorite (HTH), is a solid

which is mixed with water to form a hypochlorite solution. Calcium hypochlorite is 65-70%

concentrated.

•

Hypochlorites decompose in strength over time while in storage. Temperature, light, and physical

energy can all break down hypochlorites before they are able to react with pathogens in water.

18.2.4 Chlorine reactions related to disinfection

Chlorine reacts with water to form hypochlorous and hydrochloric acids

Cl₂

H₂O

⇐⇒

HOCl

HCl

chlorine

water

hypochlorous acid hydrochloric acid

• Hypochlorous acid dissociates in water to form the hydrogen and hypochlorite ions

HOCl

⇐⇒

H⁺

OCl⁻

hypochlorous acid

hydrogen ion

hypochlorite ion

– Hypochlorous acid is the most effective form of chlorine available to kill microorganisms

– Hypochlorite ions is much less efﬁcient disinfectant

•

The concentration of hypochlorous acid and hypochlorite ions in chlorinated water will depend on

the water’s pH

18.2 Chlorination

269

–

A higher pH facilitates the formation of more hypochlorite ions and results in less hypochlor-

ous acid in the water

•

A signiﬁcant percentage of the chlorine is still in the form of hypochlorous acid even between pH 8

and pH 9

18.2.5 Chlorine disinfection process

•

When chlorine is added to a wastewater ﬂow, it will ﬁrst react or combine with certain organic and

inorganic substances present, prior to acting on pathogens. The amount of chlorine used up as part

of these reactions is referred to as the chlorine demand

•

The free chlorine remaining after the chlorine demand is satisﬁed, is the strongest form of chlorine

available for disinfection.

Chlorine combined with ammonia (as chloramines) and organic compounds (as chloroorganic

compounds), known as combined chlorine also exhibit disinfecting properties - albeit weaker than

the free chlorine.

• Total residual chlorine is the sum of free chlorine and combined chlorine and it is the residual

chlorine concentration which represents the amount of chlorine available for disinfection

Chlorine Demand = Applied Chlorine Dose - Chlorine Residual

18.2.6 Factors affecting chlorine disinfection

The disinfection efﬁciency of chlorine depends on the following factors:

•

pH: Disinfection is more efﬁcient at a low pH when large quantities of hypochlorous acid are

present than at a high pH when hypochlorite ions is the dominant species in the water

Concentration: Contact Time Ratio (CT): For effective chlorine disinfection both sufﬁcient chlorine

dosages – concentration (C) as well as contact time (T) are necessary. Generally both of these

factors must be worked out experimentally for a given system

•

Disinfection activity can be expressed as the product of disinfection concentration (C) and contact

time (T) - CT

• The same CT values will achieve the same amount of inactivation

•

Temperature: Colder temperatures are less favorable for disinfection. Proper contacting or mix-

ing or agitation: This is necessary to make sure that the chlorine applied contacts or reaches the

microbial cells

Organic and inorganic material present: The chlorine used by these organic and inorganic reducing

substances including metal ions, organic matter and ammonia, is deﬁned as the chlorine demand.

So that the amount of chlorine that has to be added to wastewater for different purposes will also

vary.

18.2.7 Chlorine Application

•

Continuous chlorination of systems less than 75 gpm is by the use of a hypochlorinator where a

motor driven pump pulls the hypochlorite solution out of a holding chamber and pumps it into

the water to be treated. Where the pipe from the pump joins the pipe carrying the raw water, the

Venturi effect creates a small vacuum and pulls the chlorine solution into the water.

For larger system, chlorinators - devices which introduce chlorine gas to water using liquid chlorine

supplied in steel cylinders are used.

•

• In the commonly used Vacuum Chlorinator:

270

Chapter 18. Disinfection

–

Chlorine gas is pulled from the cylinder into the source water by a vacuum created by water

ﬂowing through the injector and creating a negative head.

This negative head forces open the pressure regulating valve on the cylinder and allows

chlorine gas to ﬂow out of the cylinder and into the chlorinator.

Once the gas has entered the chlorinator, the chlorine feed rate is measured using an indicator

known as a rotameter

Just beyond the rotameter, the chlorine gas ﬂows past a regulating device (a V-notch plug or a

valve) which is used to adjust the chlorine feed rate.

Figure 18.1: Vacuum Chlorinator

18.2.8 Chlorination Byproducts

•

Main drawback of chlorine disinfection is the adverse health effects of the byproducts - Disinfec-

tion by-products ( DBPs ) formed from its reaction with certain organic compounds present in the

water.

•

Adverse health effects on humans exposed to DBPs through drinking-water and oral, dermal, and

inhalational contact with chlorinated water, include cancers of vital organs.

Halogenated trihalomethanes ( THMs ) and haloacetic acids ( HAAs ) are two major classes of

disinfection byproducts (DBPs) commonly found in waters disinfected with chlorine.

At the present time, about 90% of U.S. water utilities use chlorine to disinfect water. Although

chlorine has virtually eliminated the risks of waterborne disease such as typhoid fever, cholera, and

dysentery, recent studies have shown risks associated with byproducts of chlorine—a reason why

water utilities already have been looking at alternative methods for disinfecting water.

• Approaches for reducing DBPs includes:

–

Avoiding pre-chlorination where chlorine is added to the raw water before coagulation and

ﬁltration.

– Removal of organics using Aeration or adsorption on activated carbon

–

reevaluating the chlorine dosing to a level which will accomplish the same degree of disinfec-

tion with a lower chlorine dosage.

– Another current approach is using alternative disinfection methods.

18.2 Chlorination

271

18.2.9 Chloroamination

•

When chlorine is added to water containing ammonia, chlorine reacts with ammonia to form chlo-

ramines.

•

Chloramines has disinfection properties and although it is a weaker disinfectant than chlorine,

it is more stable enabling to extend its disinfectant beneﬁts through a wider range of the water

distribution system.

•

Additionally, water disinfection using chloramines provides beneﬁt related to fewer taste and odor

issues when compared to water disinfected with chlorine.

Water utilities practice chloroamination - practice of utilizing the disinfectant properties of chlo-

ramines, by feeding chlorine to the water containing ammonia.

• When chlorine is added to water containing ammonia, the :

– an initial absence of any increase in combined chlorine residual

– followed by a decrease in the combined chlorine residual along with ammonia concentrations

–

followed by an increase in free chlorine residual and near complete removal of ammonia as

nitrogen gas.

– Chlorine reacts with ammonia to form chloramines

1. First the free chlorine in contact with ammonia forms monochloramine and water

Monochloramine has disinfection properties

Dominates when Cl:N mass ratio is 0 to 5:1

The breakpoint curve rises at about 1:1 during monochloramine formation

NH₃+HOCl → NH₂Cl(monochloramine)+H2O

2. Monochloramine reacts further with chlorine to give dichloramine and water

HOCl +NH₂Cl → NHCl₂+H₂O

Also, monochloramine auto decomposes into dichloramine

2NH₂Cl → NHCl₂+NH₃

Between dichloramine is formed between 5:1 and 7:1 Cl:N mass ratio

When you are getting signiﬁcant dichloramine, the breakpoint curve will start

dropping

NH₂Cl +HOCl → NHCl₂(dichloramine)

at pH 7.5, monochloramine is the dominant chloramine species as pH decreases from

7.5, dichloramine becomes the dominant chloramine species increases in the chlorine to

nitrogen dose ratio results in corresponding increases of nitrogen trichloride, but only

when the pH is < 7.4

3. Formation of nitrogen trichloride from the reaction of chlorine and dichloramine does

not typically occur as it is the favored product at low pH - <4

NHCl₂+HOCl → NCl₃(nitrogentrichloride)

4. Additional free chlorine +chloramines → H⁺+H2O+N₂

–

Chloramine levels up to 4 milligrams per liter (mg/L) or 4 parts per million (ppm) are consid-

ered safe in drinking water. At these levels, harmful health effects are unlikely to occur.

Chloramines have disinfection properties albeit much lower than free chlorine ( 5% of free

available chlorine) but last much longer in the system than free chlorine.

Monochloramine is about 2,000 and 100,000 times less effective than free chlorine for the

inactivation of E. Coli and rotaviruses, respectively.

After the breakpoint, free chlorine residuals develop. Free chlorine residuals usually destroy

odors, kill microorganisms and oxidize organic matter.

272

Chapter 18. Disinfection

–

Breakpoint chlorination is the application of sufﬁcient chlorine beyond the chlorine demand

to maintain a free available chlorine residual.

Theoretically chlorine requirement = Wt. NH₃-N x 7.6

in practice (Margin of safety) = Wt. NH₃-N x 10

–

Thus, breakpoint chlorination is possible if ratio of Cl₂to ammonia exceeds 10:1 then free

Cl₂may exist. Cl₂demand will be high because the reaction of free Cl₂with nitrite and other

organic compounds.

Theoretically, while microorganisms are killed as the chlorine demand is being satisﬁed,

disinfection is generally the result of chlorine residual or the amount of chlorine remaining

after the chlorine demand has been satisﬁed.

The following chart is a graphical explaination of the concept of Breakpoint Chlorination

– Point A is at the beginning of chlorine application

–

Between Points A and B, the chlorine dosage produces no residual because of an immediate

chlorine demand caused by fast-reacting ions from metal salts and H₂S.

– Point B is the beginning of the reaction between chlorine and ammonia present

– Mono and dichloramines are formed between points B and C

–

Zone 1 - between points A and C, is the combined zone and has mono and di chloramines

and ammonia. Mono chloramines is a stable disinfectant while dichloramines is a strong

disinfectant but unstable.

–

After the maximum combined residual is reached (point C), further chlorine doses decrease

the residual due to chloramine oxidation to dichloramine, occurring between points C and D.

18.2 Chlorination

This is Zone 2 - Breakpoint Zone

273

–

Point D represents the breakpoint - the point at which chlorine demand has been satisﬁed and

additional chlorine appears as free residuals

–

Between points D and E, free available residual chlorine increases in direct proportion to the

amount of chlorine applied. This is Zone 3 which is the free chlorine zone and has hypochlor-

ous acid but no ammonia.

–

Factors that affect breakpoint chlorination are initial ammonia nitrogen concentration, pH,

temperature, and demand exerted by other inorganic and organic species

Weight ratio of chlorine applied to initial ammonia nitrogen must be 8:1 or greater for the

breakpoint to be reached. If the weight ratio is less than 8:1, there is insufﬁcient chlorine

present to oxidize the chlorinated nitrogen compounds initially formed

When instantaneous chlorine residuals are required, the chlorine needed to provide free avail-

able chlorine residuals may be 20 or more times the quantity of ammonia present. Reaction

rates are fastest at pH 7-8 and high temperatures

–

18.2.10 Chlorine dosing terms

•

Chlorine dose - the amount of chlorine added to the system. It can be determined by adding the

desired residual for the ﬁnished water to the chlorine demand of the untreated water. Dosage can be

either milligrams per liter (mg/L) or pounds per day (lb/day).

•

Chlorine Demand - the amount of chlorine consumed by iron, manganese, turbidity, algae, and

microorganisms in the water. Because the reaction between chlorine and microorganisms is not in-

stantaneous, demand is relative to time. For instance, the demand 5 minutes after applying chlorine

will be less than the demand after 20 minutes.

•

Free chlorine - free chlorine refers to all chlorine present in the water as Cl₂(g), HOCl(aq) and

OCl⁻(aq).

Combined residual - is the result of combining free chlorine with nitrogen compounds. Combined

residuals are also referred to as chloramines.

Total chlorine residual - is the mathematical combination of free chlorine and combined residuals.

Total residual can be determined directly with standard chlorine residual test kits. Residual, like

demand, is based on time. The longer the time after dosage, the lower the residual will be, until

all of the demand has been satisﬁed. Residual, like demand, is expressed in mg/L. The presence

of a free residual usually provides a high degree of assurance that the disinfection of the water is

complete.

ChlorineDose(mg/L) = ChlorineDemand+ChlorineResidual

• Demand, like dosage, is expressed in mg/L. The chlorine demand is as follows:

18.2.11 Chlorine dosing problems

Example 1:

Determine the chlorinator setting (lb/day) required to treat a ﬂow of 4MGD with a chlorine dose of

5mg/L.

Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal

Chlorine feed rate (lb/ day ) = 5mg/L×4MGD×8.34lb/gal

Chlorine feed rate (lb/ day ) = 167lb/ day

274

Chapter 18. Disinfection

Example 2 :

A pipeline that is 12 inches in diameter and 1400ft long is to be treated with a chlorine dose of 48mg/L

How many lb of chlorine will this require?

First determine the gallon volume of the pipeline:

Volume (gal) = 0.785×D²× length (ft)×7.48gal/cuft

Volume (gal) = 0.785×(1ft)²×1400ft×7.48gal/cuft Volume (gal) = 8221gal

Next calculate the amount of chlorine required:

Chlorine feed rate (lb/ day ) = Chlorine (mg/L) x Flow ( MGD) ×8.34lb/gal

Chlorine feed rate (lb/ day ) = 48mg/L×0.008221MGD×8.34lb/gal

Chlorine feed rate (lb/ day ) = 3.3lb

Example 3:

A water sample is tested and found to have a chlorine demand of 1.7mg/L. If the desired chlorine residual

is 0.9mg/L, what is the desired chlorine dose (in mg/L )?

Chlorine Dose (mg/L) = Chlorine Demand + Chlorine Residual

Chlorine Dose (mg/L) = 1.7mg/L+0.9mg/L

Chlorine Dose(mg/L) = 2.6mg/L

Example 4:

The chlorine dosage for water is 2.7mg/L. If the chlorine residual after a 30-minute contact time is found

to be 0.7mg/L, what is the chlorine demand (in mg/L )?

Chlorine Demand = Chlorine Dose − Chlorine Residual

Chlorine Demand = 2.7mg/L−0.7mg/L

Chlorine Demand = 2.0mg/L

Example 5:

What should the chlorinator seting be (lb/day) to treat a ﬂow of 2.35MGD if the chlorine demand is

3.2mg/L and a chlorine residual of 0.9mg/L is desired?

First, determine the chlorine dosage (in mg/L ):

Chlorine Dose (mg/L) = Chlorine Demand + Chlorine Residual

Chlorine Dose (mg/L) = 3.2mg/L+0.9mg/L

Chlorine Dose (mg/L) = 4.1mg/L

Next calculate the chlorine dosage (feed rate) in lb/ day:

Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal

Chlorine feed rate (lb/ day ) = 4.1mg/L×2.35MGD×8.34lb/gal

Chlorine feed rate (lb/ day ) = 80.4lb/ day

Example 6:

A chlorinator setting is increased by 2lb/ day. The chlorine residual before the increased dosage was

0.2mg/L. After the increased chlorine dose, the chlorine residual was 0.5mg/L. The average ﬂow rate

being chlorinated is 1.25MGD. Is the water being chlorinated beyond the breakpoint?

First calculate the expected increase in chlorine residual:

Chlorine feed rate (lb/ day ) = Chlorine (mg/L) x Flow (MGD)×8.34lb/gal

2lb/ day = ×mg/L×1.25MGD×8.34lb/gal

x = 2/(1.25×8.34)

x = 0.19mg/L

Actual increase in residual is:

0.5mg/L−0.19mg/L = 0.31mg/L

Example 7:

18.2 Chlorination

275

A chlorinator setting of 18lb chlorine per 24 hours result in a chlorine residual of 0.3mg/L. The chlori-

nator setting is increased to 22lb per 24 hours. The chlorine residual increased to 0.4mg/L at this new

dosage rate. The average ﬂow being treated is 1.4MGD. On the basis of these data, is the water being

chlorinated past the breakpoint?

First calculate the expected increase in chlorine residual:

Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal

4lb/ day = xmg/L×1.4MGD×8.34lb/gal

x = 4/(1.4×8.34)

x = 0.34mg/L

Next calculate the actual increase in residual:

0.4mg/L−0.3mg/L = 0.1mg/L

Chapter Assessment

1. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

3. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

4. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

5. Chlorine demand is deﬁned as the amount of chlorine remaining in the waste water at the end of a

speciﬁc contact period.

a. True

*b. False

6. Chlorine disinfection is more effective at higher pH

a. True

*b. False

7. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

8. Chlorine feed rate and chlorine residual are both usually expressed in units of ppm.

*a. True

b. False

9. Chlorine is used to sterilize wastewater in order to insure protection of public health.

278

Chapter 18. Disinfection

a. True

*b. False

10. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

11. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

12. Chlorine demand is deﬁned as the amount of chlorine remaining in the waste water at the end of a

speciﬁc contact period.

a. True

*b. False

13. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

14. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

*b. False

15. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

16. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

17. Chlorine demand is deﬁned as the amount of chlorine remaining in the waste water at the end of a

speciﬁc contact period.

a. True

*b. False

18. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

19. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

*b. False

20. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

21. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

18.2 Chlorination

279

22. Before any new or repaired chlorine piping system is placed into operation, it should be pressure

tested with water to detect any possible leaks before chlorine gas or liquid is used.

a. True

*b. False

23. To produce a savings in chlorination cost and power consumption, chlorine feed rates for odor

control should be determined at low ﬂow conditions.

*a. True

b. False

24. Chlorine is used to sterilize wastewater in order to insure protection of public health.

a. True

*b. False

25. The purpose of disinfection with chlorine is to destroy pathogenic organisms.

*a. True

b. False

26. Oxidation pond efﬂuents are easily disinfected with chlorine because of the large amount of algae

present

a. True

*b. False

27. The chlorine demand of wastewater equals the chlorine dosage plus the residual.

a. True

*b. False

28. Post chlorination is employed primarily for odor control and BOD reduction.

a. True

*b. False

29. The safety plugs on a one-ton chlorine tank are designed to soften or melt at temperatures in excess

of 158°F.

*a. True

b. False

30. When moving a chlorine cylinder a short distance, for example 10 feet, it is not necessary to replace

the protective cap.

a. True

*b. False

31. A leaking chlorine cylinder cannot be transported.

*a. True

b. False

32. Canister-type gas masks are not recommended for use when working ·on chlorine leaks, because

they do not supply oxygen to the wearer.

*a. True

b. False

33. Canister-type gas masks may be used to attend chlorine gas leaks at concentrations in the range of

two to ﬁve percent.

a. True

*b. False

34. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

280

Chapter 18. Disinfection

*b. False

35. Objective of the disinfection process is to sterilize the wastewater

a. True

*b. False

36. Caution needs to be exercised when handling chlorine due to its explosive properties

a. True

*b. False

37. It is important that the (PPE) including self-contained breathing apparatus, for handling leaks from

chlorine cylinders is located inside the chlorine storage building to ensure easy access

a. True

*b. False

38. Ammonia solution is used for ﬁnding leaks in chlorine cylinders

*a. True

b. False

39. Sodium hypochlorite is more effective and less expensive than chlorine gas

a. True

*b. False

40. Sodium hypochlorite is more effective and less expensive than chlorine gas

a. True

*b. False

41. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

42. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

43. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

44. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

45. Chlorine demand is deﬁned as the amount of chlorine remaining in the waste water at the end of a

speciﬁc contact period.

a. True

*b. False

46. Chlorine disinfection is more effective at higher pH

a. True

*b. False

47. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

48. Chlorine feed rate and chlorine residual are both usually expressed in units of ppm.

18.2 Chlorination

281

*a. True

b. False

49. Chlorine is used to sterilize wastewater in order to insure protection of public health.

a. True

*b. False

50. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

51. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

52. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

53. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

*b. False

54. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

55. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

*b. False

56. Sodium hypochlorite is more effective and less expensive than chlorine gas

a. True

*b. False

57. The chlorine dose rate plus the chlorine residual equals the chlorine demand.

a. True

*b. False

58. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air

a. True

*b. False

59. Chlorine demand is deﬁned as the amount of chlorine remaining in the waste water at the end of a

speciﬁc contact period.

a. True

*b. False

60. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the

amount of residual chlorine remaining after a given contact time.

a. True

*b. False

61. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.

a. True

*b. False

282

Chapter 18. Disinfection

62. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports

even when empty. Except when actually being moved to or from storage.

*a. True

b. False

63. Chlorine demand is a good indicator of effective disinfection

a. True

*b. False

64. A chlorine cylinder valve is thought to be leaking. If ammonia vapor is passed over the valve, the

presence of a leak is indicated by

a. A hissing noise.

*b. A white cloud.

c. An odor of hydrogen sulﬁde.

d. Red smoke

65. A chlorine residual is often maintained in a plant efﬂuent:

a. to keep the chlorinator working.

b. for control of ﬂuctuation of wastewater ﬂow.

c. for testing purposes.

*d. to protect the bacteriological quality of the receiving water.

e. None of the above.

66. Acids should never be added to chlorine solutions as they

*a. Cause chlorine gas to be released.

b. Corrode or "eat away" the solution tank.

c. Decrease the disinfecting properties of chlorine.

d. Result in the formation of a chloride precipitate. „

67. An amperometric titrater is used to measure

a. Alkalinity.

*b. Chlorine residual.

c. Conductivity.

d. COD.

68. An operator should never enter a room containing a high concentration of chlorine gas without

a. Staying low on the ﬂoor.

b. Holding breath and have help standing by.

*c. Having self-contained air or oxygen supply and help standing by.

d. Covering nose and mouth with a wet handkerchief.

69. As water temperatures decrease, the disinfecting action of chlorine

*a. Decreases.

b. Increases.

c. Remains the same.

70. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that

is in service is normally determined by:

a. knowing both the pressure and temperature of the cylinder pressure gauges.

b. rotameter readings.

*c. weighing of the cylinder or tank

d. the chlorine residual test .

e. None of the above.

18.2 Chlorination

283

71. At what level should the exhaust be drawn off in a chlorination room?

a. At chest height.

b. At least two feet above the height of the chlorine cylinder.

c. Near the ceiling.

*d. Near the ﬂoor.

72. Chloramines are

*a. Combined chlorine.

b. Enzymes.

c. Found in polluted air.

d. Free chlorine.

73. Chlorine gas

a. Is lighter than air.

*b. Is heavier than air.

c. Is pink in color.

d. Will liquify at 70 degrees F.

74. Chlorine gas is

a. Colorless.

*b. Heavier than air.

c. Non-toxic.

d. Odorless.

75. Chlorine is:

a. Colorless

b. Explosive

*c. Toxic

d. All of the above

76. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitriﬁed activated sludge

efﬂuent having a pH of 6.8 and a temperature of 67F. Ammonia-nitrogen is found to range from 2

to 3 mg/L in this efﬂuent. Disinfection in this efﬂuent might be difﬁcult because:

a. a temperature of 70’ F or higher is necessary in order to achieve effective disinfection.

b. chloramines are present most of the time.

c. the chlorine dose rate is too low.

d. the pH of this efﬂuent will limit the effectiveness of free chlorine.

*e. the ratio of chlorine to ammonia-nitrogen may make it difﬁcult at times to maintain adequate

chlorine residual.

77. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

78. Chlorine is used to

*a. Disinfect.

b. Prevent corrosion.

c. Raise the pH.

d. Stabilize organics.

284

79. Chlorine residual may be determined using the reagent

Chapter 18. Disinfection

*a. Diethyl-p-phenylenediamine (DPD).

b. Ethylendiamine tetraacetic acid (EDTA).

c. Polychlorinated biphenyls (PCB).

d. Sodium thiosulfate (Na2S203)

80. "Chlorine residual" refers to:

a. the amount of chlorine remaining in the ton cylinder after use.

b. the amount of chlorine consumed during disinfection.

*c. the chlorine remaining after disinfection.

d. the chlorine that displays no disinfection power.

e. the residue left after the evaporation of chlorine gas.

81. The effectiveness of chlorine disinfection is measured by:

a. the chlorine demand

b. the chlorine dosage

c. the total chlorine residual

*d. the coliform concentration of the efﬂuent

82. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:

a. Pressure gauges.

b. Rotometers.

*c. Weighings.

d. Chlorine residuals.

e. Ammonia equivalents.

83. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?

a. 100

b. 250

*c. 460

d. 490

84. Which of the following discharges would in general, require the lowest chlorine dosage to ensure

adequate disinfection?

a. Primary plant efﬂuent

b. Activated sludge plant efﬂuent

c. Trickling ﬁlter plant efﬂuent

*d. Sand ﬁlter efﬂuent

e. Stabilization pond efﬂuent

85. Which of the following are factors that may inﬂuence the effectiveness of chlorine?

a. Chlorine dose rate

b. Contact time

c. Suspended solids concentration of the wastewater being disinfected.

*d. Only (a) and (b)

e. (a), (b), and (c)

86. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

18.2 Chlorination

285

87. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

88. "Chlorine residual" refers to:

a. the amount of chlorine remaining in the ton cylinder after use.

b. the amount of chlorine consumed during disinfection.

*c. the chlorine remaining after disinfection.

d. the chlorine that displays no disinfection power.

e. the residue left after the evaporation of chlorine gas.

89. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

90. A chlorine residual is often maintained in a plant efﬂuent:

a. to keep the chlorinator working.

b. for control of ﬂuctuation of wastewater ﬂow.

c. for testing purposes.

*d. to protect the bacteriological quality of the receiving water.

e. None of the above.

91. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that

is in service is normally determined by:

a. knowing both the pressure and temperature of the cylinder pressure gauges.

b. rotameter readings.

*c. weighing of the cylinder or tank

d. the chlorine residual test .

e. None of the above.

92. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

93. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?

a. 100

b. 250

*c. 460

d. 490

94. Identify the incorrect statement regarding disinfection.

a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater

286

Chapter 18. Disinfection

efﬂuent

b. HTH is a dry form of calcium hypochlorite

c. Appropriate doses of chlorine may be used to control odors, control ﬁlamentous bulking in

activated sludge mixed liquor, or reduce BOD5 of wastewater

*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less

costly

95. Hypochlorite solution is used in efﬂuent disinfection because:

a. Chlorine residual determination is more stable and accurate in hypochlorite

b. Hypochlorite residuals are more resistant to nitrite interference

*c. Chlorine gas is more hazardous to store and handle

d. Hypochlorite solution is easier and less costly to ship than gas chlorine

e. Chlorine causes too many problems with disinfection efﬁciency

96. Chlorine is:

a. Colorless

b. Explosive

*c. Toxic

d. All of the above

97. The effectiveness of chlorine disinfection is measured by:

a. the chlorine demand

b. the chlorine dosage

c. the total chlorine residual

*d. the coliform concentration of the efﬂuent

98. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:

a. Pressure gauges.

b. Rotometers.

*c. Weighings.

d. Chlorine residuals.

e. Ammonia equivalents.

99. Which of the following discharges would in general, require the lowest chlorine dosage to ensure

adequate disinfection?

a. Primary plant efﬂuent

b. Activated sludge plant efﬂuent

c. Trickling ﬁlter plant efﬂuent

*d. Sand ﬁlter efﬂuent

e. Stabilization pond efﬂuent

100. Which of the following are factors that may inﬂuence the effectiveness of chlorine?

a. Chlorine dose rate

b. Contact time

c. Suspended solids concentration of the wastewater being disinfected.

*d. Only (a) and (b)

e. (a), (b), and (c)

101. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

18.2 Chlorination

d. Protect downstream users from waterborne diseases

287

102. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

103. "Chlorine residual" refers to:

a. the amount of chlorine remaining in the ton cylinder after use.

b. the amount of chlorine consumed during disinfection.

*c. the chlorine remaining after disinfection.

d. the chlorine that displays no disinfection power.

e. the residue left after the evaporation of chlorine gas.

104. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

105. A chlorine residual is often maintained in a plant efﬂuent:

a. to keep the chlorinator working.

b. for control of ﬂuctuation of wastewater ﬂow.

c. for testing purposes.

*d. to protect the bacteriological quality of the receiving water.

e. None of the above.

106. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that

is in service is normally determined by:

a. knowing both the pressure and temperature of the cylinder pressure gauges.

b. rotameter readings.

*c. weighing of the cylinder or tank

d. the chlorine residual test .

e. None of the above.

107. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

108. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?

a. 100

b. 250

*c. 460

d. 490

109. Identify the incorrect statement regarding disinfection.

288

Chapter 18. Disinfection

a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater

efﬂuent

b. HTH is a dry form of calcium hypochlorite

c. Appropriate doses of chlorine may be used to control odors, control ﬁlamentous bulking in

activated sludge mixed liquor, or reduce BOD5 of wastewater

*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less

costly

110. Hypochlorite solution is used in efﬂuent disinfection because:

a. Chlorine residual determination is more stable and accurate in hypochlorite

b. Hypochlorite residuals are more resistant to nitrite interference

*c. Chlorine gas is more hazardous to store and handle

d. Hypochlorite solution is easier and less costly to ship than gas chlorine

e. Chlorine causes too many problems with disinfection efﬁciency

111. Chlorine is:

a. Colorless

b. Explosive

*c. Toxic

d. All of the above

112. The difference between the amount of chlorine added and the amount of chlorine remaining after

the contact period is referred to as:

*a. the chlorine demand

b. free chlorine residual

c. total chlorine residual

d. combined chlorine residual

e. free available chlorine

113. The ultimate measure of the effectiveness of chlorination in disinfection is:

a. the measurement of chlorine dosage

b. meeting the chlorine demand of the wastewater

c. the establishment of a chlorine demand

*d. effective reduction of the coliform count

e. none of the above

114. A chlorine cylinder valve is thought to be leaking. If ammonia vapor is passed over the valve, the

presence of a leak is indicated by

a. A hissing noise.

*b. A white cloud.

c. An odor of hydrogen sulﬁde.

d. Red smoke

115. A chlorine residual is often maintained in a plant efﬂuent:

a. to keep the chlorinator working.

b. for control of ﬂuctuation of wastewater ﬂow.

c. for testing purposes.

*d. to protect the bacteriological quality of the receiving water.

e. None of the above.

116. Acids should never be added to chlorine solutions as they

*a. Cause chlorine gas to be released.

18.2 Chlorination

b. Corrode or "eat away" the solution tank.

289

c. Decrease the disinfecting properties of chlorine.

d. Result in the formation of a chloride precipitate. „

117. An amperometric titrater is used to measure

a. Alkalinity.

*b. Chlorine residual.

c. Conductivity.

d. COD.

118. An operator should never enter a room containing a high concentration of chlorine gas without

a. Staying low on the ﬂoor.

b. Holding breath and have help standing by.

*c. Having self-contained air or oxygen supply and help standing by.

d. Covering nose and mouth with a wet handkerchief.

119. As water temperatures decrease, the disinfecting action of chlorine

*a. Decreases.

b. Increases.

c. Remains the same.

120. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that

is in service is normally determined by:

a. knowing both the pressure and temperature of the cylinder pressure gauges.

b. rotameter readings.

*c. weighing of the cylinder or tank

d. the chlorine residual test .

e. None of the above.

121. At what level should the exhaust be drawn off in a chlorination room?

a. At chest height.

b. At least two feet above the height of the chlorine cylinder.

c. Near the ceiling.

*d. Near the ﬂoor.

122. Chloramines are

*a. Combined chlorine.

b. Enzymes.

c. Found in polluted air.

d. Free chlorine.

123. Chlorine gas

a. Is lighter than air.

*b. Is heavier than air.

c. Is pink in color.

d. Will liquify at 70 degrees F.

124. Chlorine gas is

a. Colorless.

*b. Heavier than air.

c. Non-toxic.

d. Odorless.

125. Chlorine is:

290

Chapter 18. Disinfection

a. Colorless

b. Explosive

*c. Toxic

d. All of the above

126. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitriﬁed activated sludge

efﬂuent having a pH of 6.8 and a temperature of 67F. Ammonia-nitrogen is found to range from 2

to 3 mg/L in this efﬂuent. Disinfection in this efﬂuent might be difﬁcult because:

a. a temperature of 70’ F or higher is necessary in order to achieve effective disinfection.

b. chloramines are present most of the time.

c. the chlorine dose rate is too low.

d. the pH of this efﬂuent will limit the effectiveness of free chlorine.

*e. the ratio of chlorine to ammonia-nitrogen may make it difﬁcult at times to maintain adequate

chlorine residual.

127. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

128. Chlorine is used to

*a. Disinfect.

b. Prevent corrosion.

c. Raise the pH.

d. Stabilize organics.

129. Chlorine residual may be determined using the reagent

*a. Diethyl-p-phenylenediamine (DPD).

b. Ethylendiamine tetraacetic acid (EDTA).

c. Polychlorinated biphenyls (PCB).

d. Sodium thiosulfate (Na2S203)

130. "Chlorine residual" refers to:

a. the amount of chlorine remaining in the ton cylinder after use.

b. the amount of chlorine consumed during disinfection.

*c. the chlorine remaining after disinfection.

d. the chlorine that displays no disinfection power.

e. the residue left after the evaporation of chlorine gas.

131. The effectiveness of chlorine disinfection is measured by:

a. the chlorine demand

b. the chlorine dosage

c. the total chlorine residual

*d. the coliform concentration of the efﬂuent

132. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:

a. Pressure gauges.

b. Rotometers.

*c. Weighings.

d. Chlorine residuals.

18.2 Chlorination

e. Ammonia equivalents.

291

133. Which of the following discharges would in general, require the lowest chlorine dosage to ensure

adequate disinfection?

a. Primary plant efﬂuent

b. Activated sludge plant efﬂuent

c. Trickling ﬁlter plant efﬂuent

*d. Sand ﬁlter efﬂuent

e. Stabilization pond efﬂuent

134. Which of the following are factors that may inﬂuence the effectiveness of chlorine?

a. Chlorine dose rate

b. Contact time

c. Suspended solids concentration of the wastewater being disinfected.

*d. Only (a) and (b)

e. (a), (b), and (c)

135. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

136. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

137. "Chlorine residual" refers to:

a. the amount of chlorine remaining in the ton cylinder after use.

b. the amount of chlorine consumed during disinfection.

*c. the chlorine remaining after disinfection.

d. the chlorine that displays no disinfection power.

e. the residue left after the evaporation of chlorine gas.

138. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

139. A chlorine residual is often maintained in a plant efﬂuent:

a. to keep the chlorinator working.

b. for control of ﬂuctuation of wastewater ﬂow.

c. for testing purposes.

*d. to protect the bacteriological quality of the receiving water.

e. None of the above.

292

Chapter 18. Disinfection

140. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that

is in service is normally determined by:

a. knowing both the pressure and temperature of the cylinder pressure gauges.

b. rotameter readings.

*c. weighing of the cylinder or tank

d. the chlorine residual test .

e. None of the above.

141. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

142. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?

a. 100

b. 250

*c. 460

d. 490

143. Identify the incorrect statement regarding disinfection.

a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater

efﬂuent

b. HTH is a dry form of calcium hypochlorite

c. Appropriate doses of chlorine may be used to control odors, control ﬁlamentous bulking in

activated sludge mixed liquor, or reduce BOD5 of wastewater

*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less

costly

144. Hypochlorite solution is used in efﬂuent disinfection because:

a. Chlorine residual determination is more stable and accurate in hypochlorite

b. Hypochlorite residuals are more resistant to nitrite interference

*c. Chlorine gas is more hazardous to store and handle

d. Hypochlorite solution is easier and less costly to ship than gas chlorine

e. Chlorine causes too many problems with disinfection efﬁciency

145. Chlorine is:

a. Colorless

b. Explosive

*c. Toxic

d. All of the above

146. Exhaust from a chlorinator room should be taken from

a. Anywhere–the location is not important.

*b. At ﬂoor level.

c. Close to the entrance.

d. In the ceiling.

147. How many pounds of chlorine gas is necessary to treat 4,000,000 gallons of wastewater at a dosage

of 2 mg/L?

a. 61 lbs.

b. 65 lbs.

18.2 Chlorination

293

*c. 67 lbs.

d. 69 lbs.

148. How much chlorine is needed to provide 10 mg/L dosage for a ﬂow of 2 MGD?

a. 75 lbs.

b. 83 lbs.

c. 130 lbs.

*d. 167 lbs.

149. Hypochlorite solution is used in efﬂuent disinfection because:

a. Chlorine residual determination is more stable and accurate in hypochlorite

b. Hypochlorite residuals are more resistant to nitrite interference

*c. Chlorine gas is more hazardous to store and handle

d. Hypochlorite solution is easier and less costly to ship than gas chlorine

e. Chlorine causes too many problems with disinfection efﬁciency

150. Identify the incorrect statement regarding disinfection.

a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater

efﬂuent

b. HTH is a dry form of calcium hypochlorite

c. Appropriate doses of chlorine may be used to control odors, control ﬁlamentous bulking in

activated sludge mixed liquor, or reduce BOD5 of wastewater

*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less

costly

151. If a chlorinator is connected to the bottom valve of a half full one-ton cylinder, the chlorinator is

withdrawing

a. Chlorine gas.

*b. Liquid chlorine.

c. Liquid or gas chlorine, depending upon temperature.

d. Nothing, since there is only one connection.

152. If you encounter a liquid chlorine leak in a one-ton cylinder, what action will immediately help to

reduce the effect of the leak?

a. Spray it with water.

b. Apply ice. to leaking area.

c. Spray it with ammonia solution.

*d. Rotate cylinder so leak is in uppermost position.

153. In the application of chlorine for disinfection, which of the following is not normally an operational

consideration?

a. Mixing

b. Contact time

*c. DO

d. pH

e. None of the above

154. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?

a. 100

b. 250

*c. 460

d. 490

294

Chapter 18. Disinfection

155. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:

a. Pressure gauges.

b. Rotometers.

*c. Weighings.

d. Chlorine residuals.

e. Ammonia equivalents.

156. The amperometric titration method is used to measure:

a. Alkalinity.

*b. Chlorine residual.

c. pH.

d. Total hardness.

157. The fundamental purpose of disinfection is to:

a. Destroy fecal coliform bacteria

b. Destroy all bacteria

*c. Destroy pathogenic organisms

d. Protect downstream users from waterborne diseases

158. The measure of the effectiveness of chlorine in disinfection is:

a. The chlorine demand.

b. The chlorine dosage.

*c. The chlorine residual.

d. The amount of chloramine formed.

e. The ﬁnal efﬂuent coliform concentration.

159. When replacing piping in your chlorinator, be sure to

a. Replace with steel lines only.

b. Replace with CIP only.

*c. Replace with corrosion resistant piping only.

d. No special reqirement.

160. Which of the following discharges would, in general, require the lowest chlorine dosage to ensure

adequate disinfection?

a. Primary plant efﬂuent

b. Activated sludge plant efﬂuent

c. Trickling ﬁlter plant efﬂuent

*d. Sand ﬁlter efﬂuent

e. Stabilization pond efﬂuent

161. Which of the following are factors that may inﬂuence the effectiveness of chlorine?

a. Chlorine dose rate

b. Contact time

c. Suspended solids concentration of the wastewater being disinfected.

*d. Only (a) and (b)

e. (a), (b), and (c)

162. Which of the following are safe procedures for handling chlorine cylinders?

a. Keep cylinders close to direct heat to prevent freezing. -

*b. Protective cap should always be in place when moving a cylinder.

c. Roll cylinders in a horizontal position.

d. Store cylinders on their sides.

18.2 Chlorination

295

163. Which of the following discharges would in general, require the lowest chlorine dosage to ensure

adequate disinfection?

a. Primary plant efﬂuent

b. Activated sludge plant efﬂuent

c. Trickling ﬁlter plant efﬂuent

*d. Sand ﬁlter efﬂuent

e. Stabilization pond efﬂuent

19. Odor Control

19.1 Background

Odors associated with wastewater, result from the release of the following:

1. compounds which are originally discharged into the sewer – chemicals, wastes, or

2. by products of biological and chemical reactions occurring in wastewater

Odors are generated from every phase of wastewater treatment. Odor characteristics, intensity, and volume

of foul air generated is dictated by the treatment process and the stage of wastewater treatment.

Odors are dependent on:

• Odor causing compounds present

• Exposed process surface area

•

External factors including temperature and atmospheric conditions including wind speed and

direction.

• Proximity and sensitivity of odor receptors

Key wastewater properties which dictate odors include:

•

pH and Temperature: These affect the rate of biological degradation and the rate at which the

odorous compounds are released.

•

ORP: ORP governs the rate of biodegradation activities in the wastewater. Lower (more negative)

implies anaerobic conditions.

• Dissolved/bound/available oxygen: Presence of oxygen will increase the wastewater ORP.

•

Conveyance time: Longer time due to longer distance and lower velocities will promote anaerobic

degradation in the wastewater.

19.2 Drivers for Odor Control

1. Air Quality Regulations

298

Chapter 19. Odor Control

In many instances, the need for odor control systems is driven by local and state air quality regula-

tions including regulations related to prevent public nuisance.

2. Low Odor Thresholds

Many of the compounds commonly found in wastewater such as H₂S, skatoles have perceptible

odors even at very low concentrations at parts per billion levels (ppb).

3. Worker Safety

Ventilation requirements to protect workers form health hazard associated with compounds such

as H₂S and to comply with ﬁre and explosion prevention related regulations including NFPA, will

generate foul air requiring treatment prior to discharge.

4. Corrosion Prevention Severe corrosion potential of wastewater conveyance and treatment infrastruc-

ture due to H₂S conversion by bacteria to sulfuric acid.

5. Good Neighbor Policy The location of the treatment plants and sewage conveyance systems in

urban areas necessitate treatment plants to adopt a good neighbor policy to establish expecta-

tions/limits of the treatment plant vis-a-vis odor, noise, light pollution etc. to ensure harmony

with its neighbors.

19.3 Odor Causing Compounds

Hydrogen sulﬁde(H₂S) is the most common compound which causes odors related to wastewater treat-

ment.

• Hydrogen sulﬁde is the most common wastewater origin odorous compound

•

it is produced in wastewater by the activity of sulfate reducing bacteria which live in the slime layer

in the sewer pipes.

• hydrogen sulﬁde characteristics:

– has an offensive smell

– it is highly toxic and has the potential to instantly kill

–

it is converted into highly corrosive sulfuric acid through microbiological activity in the

wastewater systems.

Thus, the prevention or treatment of odors, in particular hydrogen sulﬁde provides beneﬁts includ-

ing:

– safety

– preventing public nuisance, and

– corrosion prevention.

• Other common odor pollutants include

–

organic compounds - these are typically associated with the foul air from the preliminary,

primary and secondary treatment processes

reduced sulfur compounds including mercaptans which are typically byproducts of solids

decomposition

ammonia - the odor causing constituent of the foul air from dewatering operations associated

with anaerobic digested sludge.

19.4 Theory of Odor Control

Theory related to the control of the the main odor causing compounds is as follows:

19.4 Theory of Odor Control

299

300

Chapter 19. Odor Control

19.4.1 Hydrogen Sulﬁde

H₂S is generated in the wastewater from the biological reduction of dissolved sulfates and some of the

H₂S present in the wastewater will escape into the gas phase causing odors. Once formed, H₂S control is

accomplished using the following principles:

1. pH Control

• Amount of H₂S escaping into the gas phase is dictated by Henry’s Law and is pH dependent.

•

H₂S being a weak acid, when present in an alkaline (>7 pH) solution, will ionize to HS⁻

(bisulﬁde) and subsequently to S-(sulﬁde) ions.

• H₂S odors will not occur as long as H₂S remains in solution as HS⁻or S⁻

•

Increasing pH (or OH⁻conc.) does not destroy H₂S, but keeps it from escaping into the gas

phase, as long as an alkaline pH is maintained.

•

By adding alkaline chemicals, a majority of H₂S remains in solution in the wastewater and

the amount of odorous H₂S released in the gas phase is reduced.

2. Chemical Precipitation

H₂S can be precipitated using iron salts such as ferric or ferrous chloride.

3. Chemical Oxidation

Strong oxidants including bleach and hydrogen sulﬁde can oxidize H₂S to elemental sulfur.

19.4.2 Ammonia

•

Ammonia (NH₃) is produced from the bio degradation of nitrogenous material in wastewater

including proteins and uric acid.

NH₃is soluble in water and when in solution (liquid phase), some NH₃will escape into the gas

phase causing odors.

Amount of NH₃in the gas phase is dictated by Henry’s Law and is pH dependent NH₃is a weak

base (unlike H₂S which is a weak acid) and in the presence of H⁺, it ionizes to NH₄⁺(ammonium).

Under acidic conditions, NH₃is kept in the liquid phase of wastewater, reducing amount of odor-

ous NH₃released in the gas phase.

•

NH₃odors will not occur as long as NH₃remain in solution as NH₄+. pH adjustment only helps to

keep the ammonia in solution, it does not remove/destroy the ammonia.

19.5 Odor Control Technologies

The odor control technologies used in wastewater treatment can be classiﬁed into two categories:

1. Liquid Phase Odor Control

2. Vapor Phase Odor Control

19.6 Liquid Phase Odor Control Methods

• typically applied in the collections systems

•

goal for this treatment is to prevent nuisance odors associated with the production and release of

hydrogen sulﬁde

Liquid-phase odor control strategies include:

1. Chemical precipitation:

Iron salts react with the hydrogen sulﬁde present in the wastewater to form insoluble iron sulﬁde

precipitate

19.6 Liquid Phase Odor Control Methods

301

(a) H₂S - H₂S⁻pH Equilibrium Curve

(b) NH₃- NH⁺₄pH Equilibrium Curve

2. Nitrate addition:

Addition of nitrate salts promotes the activity of certain bacteria such as Thiobacillus denitriﬁcans,

typically present in wastewater, which oxidize reduced sulfur compounds (like H₂S) while den-

302

Chapter 19. Odor Control

19.6 Liquid Phase Odor Control Methods

303

itrifying the nitrate. Presence of nitrate, also, increases oxidation-reduction potential, inhibiting

the production of any odorous compounds such as hydrogen sulﬁde which are produced under

anaerobic conditions.

3. pH Control:

Addition of alkaline chemicals including caustic soda or magnesium hydroxide increases the pH

of the wastewater, preventing the escape of the hydrogen sulﬁde to the air phase as under alkaline

condition H₂S is present as HS⁻.

•

By raising the pH of wastewater through the addition of alkaline chemicals—caustic soda

(NaOH) or magnesium hydroxide (Mg(OH)2), minimizes the potential of releasing odorous

H₂S to the gas phase.

•

This treatment does not destroy the H₂S but reduces its potential to return back to the gas

phase as long as the alkaline pH is maintained. Caustic soda also helps remove the slime

layer in the collections piping where the anaerobic bacteria responsible for the formation of

hydrogen sulﬁde are present.

• Magnesium hydroxide Mg(OH)₂can also be used for raising the pH.

• Mg(OH)₂is used primarily because of it being environmentally safer compared to NaOH

•

Mg(OH)₂is supplied as specially formulated slurry to improve solubility, keeping the solids

in suspension and prevent solids from settling and getting deposited.

Advantage of using Mg(OH)₂

•

Poses less environmental risk than NaOH due to an accidental release because of its lower pH

and corrosivity.

304

Chapter 19. Odor Control

Figure 19.4: Speece Cone

Disadvantages of using Mg(OH)2

• Lower effectiveness range

• Not effective for higher H₂S concentrations

• Higher cost

4. Air/oxygen injection The injection of air or oxygen in the sewer conveyance systems prevents

developing anaerobic condition and thus formation of hydrogen sulﬁde. A Speece Cone allows for

the injection of pure oxygen into the sewage.

19.7 Vapor Phase Odor Control Methods

• is typically applied inside the plant

•

foul air from the treatment processes is captured and scrubbed using either a packed tower scrubber,

a carbon scrubber or a bioﬁlter.

19.7.1 Common Vapor Phase Odor Control Methods

1. Packed Tower Scrubbers Packed tower scrubbers are usually rectangular or cylindrical shaped.

• Foul air is injected from the bottom

• Chemicals are added to the recirculating water

• Recirculated liquor from the sump is sprayed from the top

•

Plastic packing media provides the surface to facilitate the transfer of pollutants from the gas

phase to the liquid phase.

The pollutants transferred to the liquid phase are chemically or biologically removed or

stabilized so it does not return back to the gas phase.

Recirculation water is wasted periodically by adding make-up water to prevent build-up of

the pollutant in the recirculation liquor.

• Demister prevents the carryover of water with the air leaving the scrubber at top.

19.7 Vapor Phase Odor Control Methods

305

Figure 19.5: Packed Tower Scrubber

• Chemical oxidants such as hydrogen peroxide (H₂O₂) and bleach (NaOCl), added to the recir-

culation water in the packed tower scrubber, removes the odor causing pollutants chemically.

•

Soluble organics are oxidized to less odorous compounds and reduced sulfur compounds are

converted to elemental sulfur.

•

Solubility is the key! Only those pollutants which are soluble in water and transferred from

the gas phase to liquid phase in the packed tower scrubber will be oxidized.

306

Chapter 19. Odor Control

Applications of Packed Tower Scrubber

• For controlling organics using oxidants:

– An oxidant such as hydrogen peroxide or bleach is added to recirculation water.

–

Organics control is typically exercised for foul air from the preliminary and secondary

treatment processes.

– Only organics soluble in water will be removed

• For controlling hydrogen sulﬁde using chemicals:

Hydrogen sulﬁde control in a packed tower scrubber using chemicals is accomplished us-

ing either oxidants such as hydrogen peroxide and bleach (which is a solution of sodium

hypoclorite in caustic soda) or alkaline chemicals such as caustic soda or bleach can be added

to recirculation water. Note: Bleach is both - oxidant and alkaline.

–

H₂S transferred from the gas phase to the liquid phase in the packed tower scrubber is

kept in the liquid phase by addition of alkaline chemicals—typically bleach (NaOCl) or

caustic soda (NaOH).

–

This treatment does not destroy the H₂S but reduces its potential to return back to the

gas phase.

Bleach solution used is typically supplied as a solution contains 12.5% active chlorine

and has a pH of about 11.5.

– Caustic soda has a pH of 14

– Bleach is advantageous over caustic for the following reasons:

Use of a lower pH bleach reduces the hardness precipitation potential thereby

leading to less media blockage and therefore less acid washing need.

Oxidizes odorous organics

• For controlling ammonia using chemicals:

–

NH₃is produced from the bio degradation of nitrogenous material in wastewater includ-

ing proteins and uric acid.

NH₃is soluble in water and when in solution (liquid phase), some NH₃will escape into

the gas phase causing odors.

Amount of NH₃in the gas phase is dictated by Henry’s Law and is pH dependent NH₃

is a weak base (unlike H₂S which is a weak acid) and in the presence of H⁺, it ionizes to

NH⁺₄(ammonium). Under acidic conditions, NH₃is kept in the liquid phase of wastewa-

ter, reducing amount of odorous NH₃released in the gas phase.

NH₃odors will not occur as long as NH₃remain in solution as NH₄+. pH adjustment

only helps to keep the ammonia in solution, it does not remove/destroy the ammonia.

An acid like sulfuric acid, can be used for lowering the pH of the packed tower recircula-

tion water to convert NH₃to NH₄+.

–

In typical wastewater odor control applications, the foul air ammonia concentration does

not warrant use of an acid.

– Solubility of ammonia in water itself is sufﬁcient to provide the control of ammonia.

• As a biotrickling ﬁlter for controlling H₂S and organics:

–

Packed tower scrubbers can be operated as bioscrubbers, also known as biotrickling

ﬁlters.

Bacteria in the slime layer on the packing and in the recirculation water consume the

pollutants.

Aerobic process promotes growth of sulfur oxidizing bacteria when treating foul air

19.7 Vapor Phase Odor Control Methods

307

with H₂S

–

When used for controlling H₂S, the pH of the recirculation water drops because of the

biological conversion of H₂S to sulfuric acid.

308

2. Carbon Scrubbers

Chapter 19. Odor Control

•

This odor scrubber utilizes activated carbon’s natural adsorptive property wherein odor caus-

ing compounds are attracted and held to its surface.

•

The carbon adsorbs pollutants from the foul air passing through the scrubber. Carbon scrub-

bers can be designed to remove speciﬁc target pollutants.

• Suitable for non-polar organic and inorganic compounds including H₂S

•

The carbon may be impregnated with an oxidant such as potassium permanganate or a al-

kaline substrate to enhance its effectiveness in treating foul air with organics and hydrogen

sulﬁde respectively..

•

When the capacity of the carbon bed is exhausted, the carbon can be cleaned/regenerated or

replaced.

Figure 19.6: Carbon Scrubber

19.7 Vapor Phase Odor Control Methods

309

Figure 19.7: Bioﬁlter

3. Bioﬁlters

Bioﬁlters utilize biological processes wherein microorganisms growing on a slime layer on

•

a packed substrate - bioﬁlter media, consume the odor causing compounds from the foul air

which dissolve in the slime layer as it passes through the bioﬁlter

• Bioﬁlters can remove hydrogen sulﬁde, organics and ammonia

• Typical bioﬁlter media includes organic or inert material such as compost or activated carbon.

• Foul air is passed through the bed from the bottom

•

Irrigation water is provided on the surface and/or the foul air stream is humidiﬁed to sustain

the biological growth

Advantages of bioﬁlters include:

• Simplicity of design and operation

• Minimal maintenance requirements

310

4. Ozonation

Chapter 19. Odor Control

•

Ozone, a strong oxidizing agent, is injected in conjunction with water in the headspace to

control odors.

•

Application involves the use of an on-site ozone generator for controlling odors at locations

such as a pump station wet well.

• Effective method to control odors in sensitive environments near residential neighborhoods.

•

The ozone based oxidation of compounds including H₂S and organics requires a very short

contact time.

Figure 19.8: Ozone Injection Unit

Chapter Assessment

1. Three systems used for vapor phase odor control in wastewater treatment

2. Explain how bioxide works to control H2S odor

3. Why do we need to control H2S in wastewater/wastewater treatment

4. Explain how bioxide (sodium nitrate) works in controlling odors

5. In your own words how is a bioﬁlter different from a carbon scrubber

6. Why is the control of hydrogen sulﬁde important in wastewater treatment

7. Explain how bioxide works to control H2S odor

8. Why do we need to control H2S in wastewater/wastewater treatment

9. What are the characteristics of hydrogen sulﬁde

10. Explain how bioxide (sodium nitrate) works in controlling odors

11. In your own words how is a bioﬁlter different from a carbon scrubber

12. Why is the control of hydrogen sulﬁde important in wastewater treatment

13. Three systems used for vapor phase odor control in wastewater treatment

14. Hydrogen sulﬁde removal from the foul air can be accomplished by scrubbing with an alkaline

solution

*a. True

b. False

15. Odor control of hydrogen sulﬁde can be accomplished by the use of which of the following agents?

a. hydrogen peroxide

b. chlorine

c. ozone

*d. all of the above

e. none of the above

16. Ferric chloride helps in odor control by:

a. Oxidizing the odor constituents

b. Destruction of microorganisms responsible for odors

312

Chapter 19. Odor Control

*c. Precipitating hydrogen sulﬁde

d. Raising the pH of the wastewater

17. Use of caustic soda in odor scrubbers is used for controlling:

*a. Hydrogen sulﬁde

b. Ammonia

c. Fouling

d. Organic compounds

18. Caustic soda is used in odor scrubbers for controlling:

*a. Hydrogen sulﬁde

b. Ammonia

c. Fouling

d. Organic compounds

19. Hydrogen sulﬁde control in the collection systems by caustic soda dosing is accomplished by:

*a. pH control

b. Chemical reaction

c. Oxidation

d. Biological control

20. High sulﬁde concentrations (either gaseous or dissolved) often cause problems in a wastewater

treatment plant’s inﬂuent structure. Pre-chlorination of a plant inﬂuent is routinely practiced to

control sulﬁdes and odors. At your plant the piping that supplies chlorine for pre-chlorination needs

replacing. It is developing cracks and is corroded. Rather than replace this piping immediately, you

are directed by your supervisor to identify alternate means of controlling these inﬂuent sulﬁdes. Do

the following:

(a) Identify THREE alternative methods of controlling sulﬁdes.

(b) Brieﬂy explain how each of these methods identiﬁed in (1) is able to control sulﬁdes (e.g

chlorine destroys sulﬁdes by chemical oxidization of these sulﬁdes.

substitute for chlorine gas and brieﬂy state why you think it is best.

Response:

a Identify THREE alternative methods of controlling sulﬁdes.

• pH control – Caustic/magnesium hydroxide addition in the collection system.

• Foul air treatment using chemical scrubbers – capture and treat the foul air

• Chemical precipitation using iron salts

Brieﬂy explain how each of these methods identiﬁed in (1) is able to control sulﬁdes (e.g

chlorine destroys sulﬁdes by chemical oxidization of these sulﬁdes.

•

pH control: Keeps the H₂S in the liquid phase. Caustic would also help in control-

ling/removing the slime layer which is responsible for odor/H₂S generation.

Foul air treatment using chemical or biological scrubbers – capture and treat the foul air

– for chemical scrubbers - methods used could be using an oxidizing agent (peroxide or

beach) or alkaline pH (using caustic) recirculation water.

•

Chemical precipitation using iron salts – iron salts chemically remove the hydrogen

sulﬁde by forming iron sulﬁde precipitate.

From the three alternative methods of controlling sulﬁdes select the one you feel is the best

substitute for chlorine gas and brieﬂy state why you think it is best.

•

Process changes to remove ammonia further through implementation of nitriﬁcation –

19.7 Vapor Phase Odor Control Methods

313

denitriﬁcation as part of the activated sludge process

• Iron sulﬁde – good control over anticipated sulﬁde levels in the plant

21. Why is H₂S control important? Describe the methods for controlling odor in both, in the sewer

collection systems and in the plant.

Response:

a Why is H₂S control important?

• safety

• preventing public nuisance, and

• corrosion prevention.

Describe the methods for controlling odor in both, in the sewer collection systems and in the

plant.

In sewer collection system:

i. pH adjustment by dosing with alkaline chemicals such as caustic soda or magnesium

hydroxide

ii. By injecting oxygen or air to prevent the formation of H₂S

iii. By dosing with a iron salt to precipitate the H₂S

iv. By adding sodium nitrate (Bioxide) to facilitate biodegradation of H₂S

In the plant:

By capturing and treating the foul air using the following methods:

i. Packed tower chemical scrubber using one of the following:

• Oxidizing chemicals such as hydrogen peroxide or bleach

• Chemicals like caustic soda which increase the pH of the recirculation water

ii. Use of biological removal based systems such as:

• Bioﬁlter

• Biotrickling ﬁlter

iii. Use of adsorbtive material such as activated carbon

22. What is UV radiation disinfection? Discuss maintenance and operational issues. Advantages and

disadvantages. What is the dosage measured in and what does the equation (MW-sec/cm2) mean?

316

Chapter 20. Wastewater Chemicals

20. Wastewater Chemicals

20.1 Wastewater treatment chemicals - by use/category

USE/CATEGORY

pH Control/

Alkalinity Supplement

PROCESS

CHEMICALS USED

Caustic soda

Odor Control

Secondary Treatment

Digestion

Magnesium hydroxide

Calcium oxide

Ammonia

Sodium carbonate

Muriatic acid

Oxidant

Odor Control

Disinfection

Chlorine

Sodium hypochlorite (NaOCl)

Calcium hypochlorite (HTH)

Hydrogen Peroxide

Advance Primary Treatment/

Chemically Enhanced Primary Treat-

ment (CEPT)/Phys-Chem

Primary Treatment

Secondary

Ferric Chloride

Anionic Polymer

Filament Control

Bleach

Cationic Polymer

Phosphorous Removal

Nitrogen Removal

Dechlorination

Primary Treatment

Iron Salts

Secondary Treatment

Alum (Precipitant)

Breakpoint Chlorination

Chlrorine

Sodium Hypochlorite

Disinfection

Sodium bisulﬁte

Sulfur dioxide

Flocculation/Solids Separation

Descaling

Sludge Dewatering

Sludge Thickening

Cationic Polymer

Odor Control Scrubber

Muriatic Acid

20.2 Wastewater treatment chemicals - by use/category

317

20.2 Wastewater treatment chemicals - by use/category

PROCESS

Collections

ACTION

Odor Control

CHEMICAL USED (ROLE)

Caustic Soda (pH control)

Magnesium Hydroxide (pH control)

Hydrogen Peroxide (Oxidant)

Sodium Nitrate (Biological Degradation)

Iron Salts (Precipitant)

Primary

CEPT

Ferric Chloride (Coagulant)

Anionic Polymer (Flocculant)

Secondary

Filament Control

Bleach

Cationic Polymer

Cationic Polymer (Flocculant)

WAS Thickening

Nutrient Removal

Tertiary Treatment

Phosphorous Removal

Iron Salts (Precipitant)

Alum (Precipitant)

Calcium Oxide

Alkalinity Supplementation

Ammonia

Sodium Carbonate

Disinfection

Chlorine/Bleach

Sodium Bisulﬁte

Sulfur Dioxide

Dechlorination

Dewatering

Flocculation

Cationic Polymer

Plant Odor Control

Foul Air Scrubbing

Hydrogen Peroxide (Oxidant)

Bleach (Oxidant)

Caustic Soda (pH Control)

Muriatic Acid (pH Control & Scrubber

Descaling)

Anaerobic Digestion

Hydrogen Sulﬁde Control

Alkalinity Supplementation

Iron Salts (Precipitant)

Calcium Oxide

Ammonia

Sodium Carbonate

20.2.1 Polymers in wastewater treatment

• Polymer use in wastewater treatment includes:

– For enhancing primary removal efﬁciencies

– For sludge thickening - to increase the solids content of the sludge feed to the digester

–

For solids dewatering - to reduce the digested solids hauling cost and to make the ﬁnal solids

product more manageable

– For ﬁlament control in activated sludge treatment

Both anionic and cationic polymers used in wastewater treatment are available in the following forms:

1. Dry Polymers: These are available in granular, ﬂake or bead form and have an active polymer as

high as 95%. Prior to use, the dry polymers have to be dissolved in water using specialized mixing

318

Chapter 20. Wastewater Chemicals

units

2. Emulsion Polymer: This water soluble version consists of water droplets dispersed in oil. They

have 25% to 50% active polymer content and require a specialized system to disperse it in water

prior to use.

3. Solution polymers: These are water soluble polymers in water. These polymers are relatively easy

to put into dilute solution. However, the lower active polymer content increases the shipping cost of

this type of polymer.

4. Cationic polymer is also available as a low cost solution type polymer - Mannich Polymer (man-

nich is a type of chemical reaction involving formalydehyde which is used for making this poly-

mer). However, it has certain drawbacks which include:

(a) Presence of formaldehyde which lends its offensive odor

(b) Higher viscosity which imposes operational challenges related to its use, and

ment.

The polymers use is primarily a function of the process stream. Each system is different and there are no

hard and fast rules regarding which products will work and therefore jar tests and pilot tests are conducted

as part of the product selection process.

20.3 Chemical dosing math problems

20.3.1 lbs chemicals needed given ﬂow and dosing rate

• Use lbs formula to calculate the lbs of chemicals required

•

Using the calculated lbs chemical required value, calculate the amount of that chemical at the

concentration available

So for example, if asked how much many gallons per day of bleach solution (SG 1.2)containing 12.5%

available chlorine is required to disinfect a 10 MGD ﬂow of water given the required chlorine dosage of 7

mg/l.

1. calculate the lbs of chlorine required using the lbs formula:

=10MGD ∗ 7

∗ 8.34 = 583.8 lbs chlorine per day

2. calculate the gallons of bleach which will provide the 583.3 lbs chlorine

Applying the lbs formula - note that 8.34 * SG will give the actual lbs/gal of bleach. If SG is not

provided, use only 8.34 lbs per gallon:

lbs bleach

day

gal

day

lbs bleach lbs chlorine

∗ 0.0125

gal lb bleach

583.3

= x

∗ 8.34∗1.2

gal

day

583.3

8.34∗1.2∗0.125

gal

day

=⇒ x

= 466

20.3.2 Chemical batching and dilution

These problems include questions such as: How much initial volume of a 4% polymer solution is needed

to make 3500 gallons of polymer at 0.25% concentration?

20.3 Chemical dosing math problems

319

•

These type of problems are solved using C*V relationship where C is the concentration and V is

the volume.

As C is expressed in weight/volume, C*V will equal to weight. The weight of the chemical will be

same before and after the dilution

If C₁is the concentration of the chemical before dilution and V₁is the volume of that initial con-

centration that is needed and C₂is the ﬁnal concentration that you want to make and V₂is the

volume that you are making of the ﬁnal concentration, C₁* V₁= C₂* V₂.

• Knowing C₁, C₂and V₂, we can calculate V₁as:

C₂∗V₂

V₁=

C₁

C_.25%∗V_.25%

0.25 ∗ 3500

V_4%

= 219gal

C_4%

Take 219 gallons of the 4% polymer and dilute to 3,500 gallons to give a 0.25% polymer solution.

Chapter Assessment

1. What is mannich polymer and what are its drawbacks.

2. Chemical requirements for the conditioning of sludge are normally based upon laboratory-scale "jar

tests" which determine the volume of chemical solution required for ﬂoc formation.

*a. True

b. False

3. Cationic polymers are high-molecular-weight organic compounds carrying a negative charge

a. True

*b. False

4. Anionic polymer is used for:

a. Thickening solids in a gravity thickener

b. Flocculating solids for dewatering

c. For odor control

*d. For enhancing solids and BOD removal in the primaries

5. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested

sludge using a belt press.

a. Cationic polymers are high molecular weight organic compounds carrying a negative charge.

b. A dry polymer is always a better choice for application in centrifuges than any liquid polymer

solution.

*c. Because of its viscosity, a Mannich polymer may be difﬁcult to pump.

d. All liquid polymer solutions are harmless and need not require the examination of their MSDS.

6. Either alum; ferric chloride; or lime may be used to remove solids from a secondary efﬂuent.

Which one of, the following statements is TRUE regarding these chemicals:

a. Typical dose rates for alum when it is applied for the removal of phosphorus from a secondary

efﬂuent are 1 to 10 mg/L.

b. Hydrated lime needs to be "slaked" prior to use.

*c. The safety precautions for handling liquid ferric chloride are the same as those for handling an

322

Chapter 20. Wastewater Chemicals

acid.

d. All of these chemicals raise the pH of the wastewater to which they are applied.

7. Ferric chloride helps in odor control by:

a. Oxidizing the odor constituents

b. Destruction of microorganisms responsible for odors

*c. Precipitating hydrogen sulﬁde

d. Raising the pH of the wastewater

8. Flocculation is best accomplished by

a. Decreasing alkalinity.

*b. Gentle agitation.

c. Increased sunlight.

9. Sodium hydroxide, (caustic soda) when used in wastewater:

Is typically applied at 1 - 10 mg/l when used to precipitate phosphorus in primary sedimentation

systems.

a. Should be treated as an acid with regard to safe handling.

b. Should be immediately diluted to 10% upon receiving.

*c. Raises the pH of the wastewater to which it is added.

d. Is added to ﬁltered efﬂuent to improve de-chlorination with sulfur dioxide.

10. Either alum; ferric chloride; or lime may be used to remove solids from a secondary efﬂuent.

Which one of, the following statements is TRUE regarding these chemicals:

a. Typical dose rates for alum when it is applied for the removal of phosphorus from a secondary

efﬂuent are 1 to 10 mg/L.

b. Hydrated lime needs to be "slaked" prior to use.

*c. The safety precautions for handling liquid ferric chloride are the same as those for handling an

acid.

d. All of these chemicals raise the pH of the wastewater to which they are applied.

11. Flocculation is best accomplished by

a. Decreasing alkalinity.

*b. Gentle agitation.

c. Increased sunlight.

d. Rapid mixing

12. Flocculation is best accomplished by:

a. Decreasing alkalinity

*b. Gentle agitation

c. Increased sunlight.

13. If a chemical costs S30 per ton, how much will it cost per year to treat a ﬂow of 1.5 MGD if the

average dose is 18 mg/L?

a. $803.

b. $110.

*c. $233.

d. $506.

Identify the correct statement regarding polymers.

14. Sodium hydroxide, (caustic soda) when used in wastewater:

Is typically applied at 1 - 10 mg/l when used to precipitate phosphorus in primary sedimentation

systems.

20.3 Chemical dosing math problems

323

a. Should be treated as an acid with regard to safe handling.

b. Should be immediately diluted to 10% upon receiving.

*c. Raises the pH of the wastewater to which it is added.

d. Is added to ﬁltered efﬂuent to improve de-chlorination with sulfur dioxide.

15. Which one of the following statement is TRUE regarding polymers?

a. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested

sludge using a belt press.

b. Cationic polymers are high-molecular-weight organic compound carrying a negative charge.

c. A dry polymer is always a better choice for application in centrifuges than any liquid polymer

solution.

*d. Because of its high pH, Mannich polymers may cause scale formation.

e. All. liquid polymer solutions are harmless and need not require the examination of its MSDS

sheet.

16. Which one of the following statement is TRUE regarding polymers?

a. Cationic polymers are high molecular weight organic compound carrying a negative charge.

b. A dry polymer is always a better choice for application in centrifuges than any liquid polymer

solution.

*c. Because of its high pH, Mannich polymers may cause scale formation.

d. All liquid polymer solutions are harmless and need not require the examination of its MSDS

sheet.

e. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested

sludge using a belt press.

21. Pumping Efﬁciencies and Costs

•

Pump is a machine used for moving water (and other ﬂuids) through a piping system and raise the

pressure of the water.

Pumping is accomplished by transforming the input energy - typically from an electric motor or

from other sources such as high-pressure air.

• The pump calculations in this section are for electrically driven rotodynamic pumps.

•

To move water, a pump will need to overcome resistance due its density, gravitational force and

friction.

• This resistance is dependent on:

–

Height the water needs to be raised. This height of the ﬂuid in a container is referred to as

head.

– Quantity of water involved

21.1 Glossary of Pump Calculations Terms

Force: In the English system force and weight are often used in the same way. The weight of the cubic

foot of water is 62.4 pounds. The force exerted on the bottom of the one foot cube is 62.4 pounds. If we

have two cubes stacked on top of one another, the force on the bottom will be 124.8 pounds.

Pressure: Pressure is a force per unit of area, pounds per square inch or pounds per square foot are com-

mon expressions of pressure.

Head: Pressure is directly related to the height of a column of ﬂuid. This height is called head or feet of

head. Pressure and feet of head head are directly related - for every one foot of head there is a pressure of

0.433 psi.

0.433 psi

ft (water column)

1 ft (water column)

2.31 psi

Thus,

or conversely

Note: This pressure/head will include the height the water pumped and also the head associated with friction

losses - energy loss because of the water moving through the pipe and ﬁttings.

326

Chapter 21. Pumping Efﬁciencies and Costs

Force

Area

Pressure =

12"

Pressure exerted by

a 1ft column of water

62.4 lb

12in x 12in

= 0.43 psi

12"

As 1ft³of water weighs 62.4 lbs

12"

The pressure at the bottom of a container is affected only by the height of water in the container and not

by the shape or the volume of the container. In the drawing below there are four containers all of different

shapes and sizes. The pressure at the bottom of each is the same.

The pressure exerted at the bottom of a tank is relative only to the head on the tank and not the volume

of water in the tank. For example, below are two tanks each containing 5000 gallons. The pressure at the

bottom of each is 22 psi. If half of the water were drained from the tanks the pressure at the bottom of the

elevated tank would be 17.3 psi while the pressure at the bottom of the standpipe would be 11 psi.

Velocity: Velocity is the speed that the water is moving along a pipe or through a basin. Velocity is

usually expressed in feet per second, ft/sec.

Flow: Flow is commonly expressed in gallons per minute (gpm) and/or cubic feet per second (cfs). There

is a relationship between gallons per minute and cubic feet per second. One cubic foot per second is equal

to 448.8 gallons per minute.

1cfs = 448.8gpm

21.1 Glossary of Pump Calculations Terms

327

Flow Equation; The basic equation for determining ﬂow is as follows:

Q = V ×A

Where:

Q = cfs ft³/sec

ꢂ

ꢃ

V = ft/sec

A = ft²

Static Pressure: Static implies a non-moving condition. The pressure measured when there is no water

moving in a line or the pump is not running is called static ³²pressure. This is the pressure represented by

the gauges on the tanks in the discussion above.

Dynamic Pressure: When water is allowed to run through a pipe and the pressure (called pressure head)

measured at various points along the way we ﬁnd that the pressure decreases the further we are from the

sources.

Headloss: The reason for this reduction in pressure is a phenomenon called headloss. Headloss is the

loss of energy (pressure) due to friction. The energy is lost as heat.

If the headloss in a certain pipe is 25 feet, it means the amount of energy required to overcome the fric-

tion in the pipe is equivalent to the amount of energy that would be required to lift this amount of water

straight in the air 25 feet.

In a pipe, the factors that contribute to headloss include the following:

• Roughness of pipe - If the roughness of a pipe were doubled the headloss would double.

• Length of pipe - If the length of the pipe were doubled the headloss would double.

• Diameter of pipe - If the diameter of a pipe were doubled the headloss would be cut in half

•

Velocity of water - If the velocity of the water in a pipe were doubled the headloss would be in-

creased by about four times. It should be apparent that velocity, more than any other single factor,

affects headloss. To double the velocity we would have to double the ﬂow in the line.

Pumping System Components and Fittings - Each type of ﬁtting has a speciﬁc headloss depending

upon the velocity of water through the ﬁtting. For instance the headloss though a check valve is

two and one quarter times greater than through a ninety degree elbow and ten times greater than the

headloss through an open gate valve.

•

Static Head: Static head is the distance between the suction and discharge water levels when the pump is

shut off.

Suction Lift: Suction lift is the distance between the suction water level and the center of the pump

impeller. This term is only used when the pump is in a suction lift condition. A pump is said to be in a

suction lift condition any time the eye (center) of the impeller is above the water being pumped.

Velocity Head: The amount of energy required to bring a ﬂuid from standstill to its velocity. For a given

quantity of ﬂow, the velocity head will vary indirectly with the pipe diameter.

Total Dynamic Head (TDH): The total energy needed to move water from the center line of a pump (eye

of the ﬁrst impeller of a lineshaft turbine) to some given elevation or to develop some given pressure. This

includes the static head, velocity head and the headloss due to friction.

328

Chapter 21. Pumping Efﬁciencies and Costs

Horsepower: Horsepower is a measurement of the amount of energy required to do work. Motors are

rated in horsepower. The horsepower of an electric motor is called brake horsepower. The horsepower

requirements of a pump are dependent on the ﬂow and the total dynamic head. 33,000 foot pounds per

minute of work is 1 horsepower.

Suction Head: Suction head is the distance between the suction water level and the center of the pump

impeller when the pump is in a suction head condition. A pump is said to be in a suction head condition

any time the eye (center) of the impeller is below the water level being pumped.

Velocity Head: Velocity head is the amount of energy required by the pump and motor to overcome

inertia and bring the water up to speed. Velocity head is often shown mathematically as V²/2 g. (

acceleration due to gravity −32.2ft/sec²).

g is the

Total Dynamic Head: Total dynamic head (TDH) is a theoretical distance. It is the static head, velocity

head and headloss required to get the water from one point to another.

The horsepower output of an electric motor is directly reﬂected to the amperage that the motor draws. Any

increase in horsepower requirements will give a corresponding increase in amperage.

Cavitation: Cavitation in pumps is the rapid creation and subsequent collapse of air bubbles occuring as

a result of the inlet pressure falling below the design inlet pressure or when the pump is operating at a ﬂow

rate higher than the design ﬂow rate. This collapse of the air bubbles typically manifests as a pinging or

crackling noise. Cavitation is undesirable because it can damage the impeller, cause noise and vibration,

and decrease pump efﬁciency.

21.2 Pumping Rate Calculations

•

For calculating volume pumped given the pump ﬂow rate: Multiply the pump ﬂow rate by the time

interval

Make sure:

– The time units - in the given time interval and in the pump ﬂow rate match

• For calculating time to pump a certain volume:

Step 1. Calculate the total volume pumped

Step 2. Divide the total volume by the pump ﬂow rate

Make sure:

– The volume units - in the volume that needs to be pumped and in the pump ﬂow rate match

–

The time unit in the pump ﬂow rate needs to be converted to the time unit that you need the

answer in

21.3 Power Requirements for Pumping

329

21.3 Power Requirements for Pumping

Where:

•

Input Hp is the input power to the motor which produces the Output Hp or Brake Hp - the

mechanical power which runs the pump.

• The ratio of Output Hp and Input Hp is the motor efﬁciency - η_m.

• The Output Hp is the input power (Brake Hp) to the pump to pump the water.

•

Water Hp is the rate of energy transferred to the water being pumped and can be calculated by the

formula:

H − Head of water (ft) ∗ Q − Flow (GPM)

3,960 (Conversion factor for converting GPM−ft to Hp)

• The ratio of Output Hp and Water Hp is the pump efﬁciency - η_p.

330

Chapter 21. Pumping Efﬁciencies and Costs

21.4 Summary of conversions and formulas

Total dynamic head = Friction head + Static head

1Hp = 0.746kW

33,000 ft −lb

1Hp =

min

1Hp = 3,960 GPM − ft

1psi = 2.31 ft

Water Hp = Flow ∗ Head

Brake Hp = Input Hp∗Motor e f ficiency

Water Hp = Brake Hp ∗ Pump e f ficiency, and

=⇒ Water Hp = Input Hp ∗ Motor e f ficiency ∗ Pump e f ficiency

PUMPS & PUMPING

Pump is a device for raising

To move water, need to overcome

This resistance is dependent on:

or moving water or any other ﬂuid.

resistance due its density,

• Height the water needs to be raised

• Quantity of water involved

gravitational force & friction.

Power needs to be delivered to the pump so it can provide energy. Power = Energy per time

Force needs to be exerted

by the pump to overcome

the resistance

Pump will need to

provided energy to raise

the water

Power Units

ft-lb

min

Understanding the concept of power:

1Hp = 0.746 kW

Energy = resistance force

Energy = Force x Distance

Force is the head which is measured in terms

of the height of water - inches or feet

A 150lb person climbing 50ft will expend

7500 ft-lb of work (energy)

Force = Mass x Acceleration

Watt determined that one horse on an

average could lift 330lbs 100ft in one minute

(lb_f)

(lb_m)

Energy units

ft-lb

KWh

Calories

Hp-h

1) Power requirement for climbing this in 5 minutes

lb_f= lb_m

(per deﬁnition)

33,000 ft-lb

1Hp =

min

ft-lb

min

7500

ft-lb

min

= 1500

1'=12"

8.34lb

As 1 GPM (Water) =

min

= 0.045 Hp

force

area

1'=12"

Pressure =

GPM

8.34

min

1'=12"

2) Power requirement for climbing this in 1 minute

33000 ft-lb

min

33000 ft x GPM

8.34 min

1Hp =

7500

ft-lb

min

= 0.23 Hp

Pressure exerted by

a 1ft column of water 12 in x 12 in

62.4 lb

= 0.43 psi

1Hp = 3,960 GPM-ft

so 0.43 psi = 1ft (water column)

or 1 psi = 2.3ft

1Hp is needed to raise one gallon of water 3,960

ft in one minute

Flow * Head

Friction Head

Head associated with energy

losses due energy losses due

to friction

Input Hp

Output Hp

Brake Hp

Water Hp

Eff:

Head

Total Dynamic Head

Total equivalent height

of water column

Static Head

needed to pump

Head associated with the height

the water has to be raised

Given

Calculate

What is the required water horsepower to pump 300 GPM if the suction head and discharge heads are

constant at 20 ft and 120 ft respectively and the system head losses are 5 ft.

Calculating Static Head

Solution:

300 GPM x (120-20+5 ft)

7.95Hp

GPM-ft

3960

Final Water Surface

El = y

Given

Calculate

Find the input Hp Given the water Hp equals 8Hp and the pump and motor efﬁciencies are 80%

and 50% respectively.

Solution:

Step 1: Find

knowing

Step 2: Find

knowing

Brake Hp

8Hp

Brake Hp

= Brake Hp = 16Hp

0.8 =

0.5 =

Input Hp

0.8

Input Hp =

20Hp

El = x

El = 0

El = -x

Suction

lift (x)

Given

and existing

& replacement

Calculate

savings

Suction

Head (x)

An older motor which is 82% efﬁcient is to be replaced by a new, 94% efﬁcient motor. Calculate the

annual savings for the new motor given the output horsepower from both motors is 180Hp and the elec-

tricity cost is $0.075/kWh & the motor operates 24 hours per day throughout the year.

Case A:

Initial water level

above pump

Case B:

Initial water level below

pump

180

0.94

180

0.82

Hp input to

new motor

Hp input to

old motor

= 191.5Hp

= 218.5Hp

$0.075

Kwh

$23,779

Year

365 x 24 hrs

Annual

Savings

0.746 Kw

= y-x

= y-(-x)

= y+x

Static Head

(218.5-191.5)Hp

Year

“Demand Charge” ($/kw) is imposed on consumer by the utility company to compensate it for the design & upkeep of equipment to meet the peak power draw by the consumer. The demand charge is in addition to the regular energy

consumption charge ($/kw). The demand charge is based upon highest power consumption over any 15 minute period during the billing cycle. Oversized and inefﬁcient equipment (motors & pumps) would mean higher demand charges.

21.5 Practice Problems

333

21.5 Practice Problems

1. A reservoir is 40 feet tall. Find the pressure at the bottom of the reservoir.

40ft×0.433psi/ft = 17.3psi

2. Find the height of water in a tank if the pressure at the bottom of the tank is 12 psi.

12psi÷0.433psi/ft = 27.7ft

3. If a pump discharge pressure gauge read 10 psi, the height of the water corresponding to this pres-

sure would be:

2.31 ft

10 psi×

= 23.1 ft

psi

4. A pump is set to pump 5 minutes each hour. It pumps at the rate of 35 gpm. How many gallons of

water are pumped each day?

Solution:

35 gal sludge 5 min 24 hꢀr

4,200 gallons

ꢀ

∗

min

hꢀr

ꢀ

day

Example 2: A pump operates 5 minutes each 15 minute interval. If the pump capacity is 60 gpm,

how many gallons are pumped daily?

H¨

60 gal sludge 5 min

min

28,800 gal sludge

¨H

∗

∗1440

H¨

min

¨H

15 min

day

5. Given the tank is 10ft wide, 12 ft long and 18 ft deep tank including 2 ft of freeboard when ﬁlled to

capacity. How much time (minutes) will be required to pump down this tank to a depth of 2 ft when

the tank is at maximum capacity using a 600 GPM pump

Solution:

2’ Freeboard

16’ Water Depth (Initial)

2’ Water Depth (Final)

10’ Wide

12’ Long

Volume to be pumped=12 ft ∗10 ft ∗(16−2) ft = 1,680ft³

ꢀ

ft³

gal

ꢀ

1,680ft ∗7.48

=⇒

= 21min

ꢀ

gal

ꢀ

600

min

6. 1 MGD is pumped against a 14’ head. What is the water Hp? The pump mechanical efﬁciency is

85%. What is the brake horsepower?

water Hp = ﬂow * head

1,000,000 gal

day

1440 min

∗

∗14 ft ∗

= Water Hp = 2.46 Hp

day

3,960 GPM − ft

334

Chapter 21. Pumping Efﬁciencies and Costs

pump Hp = brake Hp * pump efﬁciency

2.46

0.85

Brake Hp =

= Brake Hp = 2.89Hp

7. A 8 ft diameter cylindrical wetwell receives an average incoming ﬂow if 135 gpm and is pumped

down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-

trolled using two ﬂoats; a stop ﬂoat located at 2.5 ft and a start ﬂoat located at 16 ft. If the pump

motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this

pump if power costs are $0.11/Kwh?

When the pump is on, the volume of wetwell that will be pumped down with the 450 gpm pump

and a 135 gpm ﬂow to the wetwell:

450 gal 135 gal

315 gal

min

−

min

Minutes required to pump down the wetwell :

7.48 gal

min

315 gal

0.785∗8²∗(16−2.5) ft³∗

∗

= 16.1 min

ft³

Time to ﬁll wetwell with pump off @135gal/min inﬂuent ﬂow:

7.48 gal

min

135 gal

[0.785∗8²∗(16−2.5)] ft³∗

∗

= 37.6min

ft³

# of cycles per day:

cycle

(16.1+37.6) min

1440 min 26.8 cycles

∗

day

# of hrs pump operational:

16.1 min 26.8 cycles

hrs

60 min

7.19 hours

∗

cycle

day

Monthly electrical cost:

450 gpm∗120 ft

0.746 kW 7.19hrs 30 days $0.11

$356

month

∗

0.88∗0.77

3,960 gpm− ft

day

month

kWh

8. A 6-year old pump motor is to be replaced at a net cost of $15,800. The new motor, just like the old

one, would run 65% of the time. Both existing and replacement motors would operate at 125 output

Hp. The existing motor efﬁciency is 86% while the replacement motor would be guaranteed at 94%

efﬁciency. Electricity currently averages $0.088 per kWh.

(a) Calculate the energy cost savings per year (to the nearest dollar) if the existing motor is re-

placed with the new motor (neglect any consideration of impact upon demand charges or interest

on capital).

(b) What is payback period to the nearest tenth of a year.

Solution:

21.5 Practice Problems

335

Calculate energy cost savings per year:

125

Input Hp for old motor:

= 145.35Hp

0.86

125

0.94

Input Hp for old motor:

= 132.98Hp

Energy cost savings:

0.746 kW (365∗24∗0.65)hrs $0.088

$4,624

(145.35−132.98)Hp∗

∗

kWh

Calculate payback:

$15,800∗

= 3.4yr

$4,623.94

9. A ﬂow of 200 gpm is pumped against a total head of 4.0 feet. The pump is 78% efﬁcient and the

motor’ is 90% efﬁcient. Calculate the input Hp.

water Hp = ﬂow * head

200GPM ∗4 ft ∗

= 0.2Hp

3,960GPM − ft

water Hp=brake Hp*pump efﬁciency, and

brake Hp=input Hp*motor efﬁciency

Therefore, water Hp=input Hp*motor efﬁciency*pump efﬁciency

water Hp

0.2

input Hp=

= 0.28Hp

motor e f ficiency∗ pump e f ficiency 0.9∗0.78

Chapter Assessment

1. A 240 volt motor runs an average of 8 hours a day. If the electric meter registered 6,450 kilowatt

hours for a 31-day month, what is the motor horsepower? [35 Hp]

2. A pump is equipped with a pressure gauge in the discharge pipe that reads 100 psi. The total dis-

charge head in feet would be?

2.31ft water

100psi∗

= 231 ft water

psi

3. 900 GPM pump is pumped against a 12 ft head. What is the water Hp water Hp = ﬂow * head

900GPM ∗12 ft ∗ _{3,960GPM−ft}= 2.7Hp

4. A 50 ft³/sec ﬂow is pumped against a head of 8 feet. What is the water Hp

water Hp = ﬂow * head

7.48gal

60sec

min

ft³sec∗8 ft ∗

∗

= 45.4Hp

3,960GPM−ft

5. 1 MGD is pumped against a 14’ head. What is the water Hp? The pump mechanical efﬁciency is

85%. What is the brake horsepower?

water Hp = ﬂow * head

1,000,000gal

day

∗

_1440min∗14 ft ∗ _{3,960GPM−ft}= Water Hp = 2.46Hp

day

pump Hp = brake Hp * pump efﬁciency

2.46

0.85

brake Hp =

= Brake Hp = 2.89Hp

6. A ﬂow of 2.5 MGD is being lifted 10 feet and then pumped up another 120 feet to a storage reser-

voir. Calculate the pump output power required to lift this water. Ignore friction losses.

water Hp = ﬂow * head

2,500,000gal

day

∗

_1440min∗(120+10) ft ∗ _{3,960GPM−ft}= Water Hp = 57Hp

day

338

Chapter 21. Pumping Efﬁciencies and Costs

7. A pump is pumping 400 gpm. The suction pressure gauge indicates a pressure of 5 ft and the pump

discharge pressure gauge indicates a pressure of 100 ft. If the pump brake horse power is 12 hp,

what is the pump efﬁciency

water Hp = ﬂow * head

400gpm∗(100−5) ft ∗ _{3,960gpm−ft}= 9.6Hp

pump efﬁciency - η_p=^9.6Hp∗100 = 79%

12Hp

8. A ﬂow of 200 gpm is pumped against a total head of 4.0 feet. · The pump is 78% efﬁcient and the

motor’ is 90% efﬁcient. Calculate the input Hp.

water Hp = ﬂow * head

200GPM ∗4 ft ∗ _{3,960GPM−ft}= 0.2Hp

water Hp=brake Hp*pump efﬁciency, and

brake Hp=input Hp*motor efﬁciency

Therefore, water Hp=input Hp*motor efﬁciency*pump efﬁciency

water Hp

e f ficiency∗pump e f ficiency

0.2

0.9∗0.78

input Hp=_motor

= 0.28Hp

21.5 Practice Problems

339

9. 500,000 gpd of secondary· efﬂuent is pumped to a storage pond for reuse as golf course irrigation

water. The water is lifted 12 feet in the plant, and then pumped up another 75 feet to the storage

pond. Friction losses are assumed to be 10% of the static head. Assuming the pump efﬁciency of

70% and a motor efﬁciency of 92% and an electrical cost of $0.0725 per KWh, calculate the daily

cost of pumping this water.

Solution:

water Hp = ﬂow * head

500,000gal

day

∗

_1440min∗(87 ft −static head +87∗0.1 ft − friction head)∗ _{3,960GPM−ft}

day

= 8.39−water Hp

water Hp

e f ficiency∗pump e f ficiency

8.39

0.92∗0.70

input Hp=_motor

= 13Hp

$16.87

day

0.746kW 24hrs $0.0725

Electrical cost=13Hp∗

∗

kWh

day

10. A pump motor (93% efﬁcient) generates an output of 130 HP and runs 75% of the time. Electricity

costs an average of 8.455 cents per kilowatt-hour. What is the monthly cost of operating this pump

in $ per month?

$4,761

month

130Hp

0.93

30days

month

0.746kW 24hrs

$0.08455

kWh

∗

∗0.75∗

day

340

Chapter 21. Pumping Efﬁciencies and Costs

11. A wet well is 8 ft x 8 ft x 16.5 ft deep and receives a continuous ﬂow of 310,000 gpd. A 500 gpm

pump draws down 12 feet of water each pumping cycle. The motor that drives the pump draws 52.5

Hp when it pumps. The cost of electricity is $0.0755 per kilowatt - hour. Calculate

(a) the time it takes to pump down the wet well, and

(b) The daily electrical energy cost for this pump.

Solution:

310,000gal

day

215.3gal

min

∗

1440min

day

Volume of wetwell that will be pumped down with the 500 gpm pump and a 215.3 gpm ﬂow to the

wetwell:

500gal

min

215.3gal

min

284.7gal

min

−

Minutes required to pump down the wetwell :

7.48gal

min

8∗8∗12 ft³∗

∗

= 20.2min

284.7gal

Time to ﬁll wetwell with pump off @215.3gal/min inﬂuent ﬂow:

7.48gal

min

8∗8∗12 ft³∗

∗

= 26.7min

215.3gal

# of cycles per day:

cycle

30.7cycles

1440min

∗

day

(20.2+26.7)min

# of hrs pump operational:

30.7cycles

20.2min

hrs

10.33hours

∗

60min

cycle

day

Daily electrical cost:

$30.54

day

0.746kW 10.33hrs $0.0755

52.5Hp∗

∗

kWh

day

21.5 Practice Problems

341

12. A 6-year old pump motor is to be replaced at a net cost of $15,800. The new motor, just like the old

one, would run 65% of the time. Both existing and replacement motors would operate at 125 output

Hp. The existing motor efﬁciency is 86% while the replacement motor would be guaranteed at 94%

efﬁciency. Electricity currently averages $0.088 per kWh.

(a) Calculate the energy cost savings per year (to the nearest dollar) if the existing motor is re-

placed with the new motor (neglect any consideration of impact upon demand charges or interest

on capital).

(b) What is payback period to the nearest tenth of a year. Energy cost savings per year:

125

Input Hp for old motor:

Energy cost savings:

= 145.35Hp

= 132.98Hp

0.86

125

0.94

$4,623.94

(365∗24∗0.65)hrs

0.746kW

$0.088

(145.35−132.98)Hp∗

∗

kWh

Calculate payback:

$15,800∗ _$4,623.94= 3.4yr

13. A 8 ft diameter cylindrical wetwell receives an average incoming ﬂow if 135 gpm and is pumped

down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-

trolled using two ﬂoats; a stop ﬂoat located at 2.5 ft and a start ﬂoat located at 16 ft. If the pump

motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this

pump if power costs are $0.11/Kwh? (Ans: $356/month)

Volume of wetwell that will be pumped down with the 450 gpm pump and a 135 gpm ﬂow to the

wetwell:

450gal

min

135gal

min

315gal

−

min

Minutes required to pump down the wetwell :

7.48gal

min

0.785∗8²∗(16−2.5) ft³∗

∗

= 16.1min

315gal

Time to ﬁll wetwell with pump off @135gal/min inﬂuent ﬂow:

7.48gal

min

135gal

[0.785∗8²∗(16−2.5)] ft³∗

∗

= 37.6min

# of cycles per day:

cycle

26.8cycles

1440min

∗

day

(16.1+37.6)min

# of hrs pump operational:

26.8cycles

16.1min

hrs

7.19hours

day

∗

cycle

day

60min

Daily electrical cost:

$356

day

450gpm∗120 ft

0.88∗0.77

30days

month

0.746kW 7.19hrs

$0.11

∗

kWh

3,960gpm− ft

day

14. A pump operating at 80% efﬁeciency generates an water Hp of 60 HP and runs 75% of the time.

Assuming the pump motor is 90% efﬁcent and electricity costs an average of $0.0821 per kilowatt-

hour. The monthly (30 days) cost of operating this pump is:

342

Chapter 21. Pumping Efﬁciencies and Costs

$2,756

month

60Hp

0.90∗0.80

30days

month

0.746kW 24hrs

$0.0821

kWh

∗

∗0.75∗

−Correct Answer

day

$2,480

month

60Hp

0.8

30days

0.746kW 24hrs

$0.0821

∗

∗0.75∗

day

month

kWh

$2,205

month

60Hp

0.90

30days

month

0.746kW 24hrs

$0.0821

kWh

∗

day

$3,675

month

60Hp

0.90∗0.80

30days

month

0.746kW 24hrs

$0.0821

kWh

∗

day

15. A 8 ft diameter cylindrical wetwell receives an average incoming ﬂow if 135 gpm and is pumped

down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-

trolled using two ﬂoats; a stop ﬂoat located at 2.5 ft and a start ﬂoat located at 16 ft. If the pump

motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this

pump if power costs are $0.11/Kwh? (Ans: $356/month)

16. A 8 ft diameter cylindrical wetwell receives an average incoming ﬂow if 135 gpm and is pumped

down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-

trolled using two ﬂoats; a stop ﬂoat located at 2.5 ft and a start ﬂoat located at 16 ft. If the pump

motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this

pump if power costs are $0.11/Kwh? (Ans: $356/month)

17. If a wasting pump has a ﬁxed pump rate of 250 GPM, and your calculation indicates you must

waste 126,000 gallons, what hourly cycle rate do you set the timer?

a. Turn pump on 21 minutes every day

b. Turn pump on 504 minutes every hour

c. Turn pump on 42 minutes every day

d. Turn pump on 21 minutes every hour

Solution:

21 min

126,000 gal

day

min

24 hrs 250 gal

∗

day

18. A pump operating at 80% efﬁciency generates an water Hp of 60 HP and runs 75% of the time.

Assuming the pump motor is 90% efﬁcent and electricity costs an average of $0.0821 per kWh.

The monthly (30 days) cost of operating this pump is:

*a. $2,756

b. $2,480

c. $2,205

d. $3,675

19. How long will it take to pump down 25 feet of water in a 110 ft diameter cylindrical tank when

using a 1420 gpm pump.

a. 26 hours and 56 minutes

*b. 26 hours and 33 minutes

c. 2 hours and 47 minutes

d. 12 hours and 36 minutes

20. How long will it take to pump down 25 feet of water in a 110 ft diameter cylindrical tank when

using a 1420 gpm pump.

a. 26 hours and 56 minutes

*b. 26 hours and 33 minutes

c. 2 hours and 47 minutes

21.5 Practice Problems

343

d. 12 hours and 36 minutes

21. A positive displacement pump should be started up with the discharge valve closed in order to

avoid any problems with "air lock".

a. True

*b. False

22. Brake horse power is the input power to the motor

a. True

*b. False

23. Cost of electrical usage for pumps is based upon kilowatt per hour

a. True

*b. False

24. A centrifugal pump can be used to pump sludge.

*a. True

b. False

25. Variable speed sludge pumps may be used to keep the density of the sludge nearly constant.

*a. True

b. False

The keeping of records of plant operation and maintenance, even when a portion of the plant is

temporarily out of balance, is an integral part of good operation.

*a. True

b. False

26. A positive displacement pump could be damaged if it is started with the discharge valve closed.

*a. True

b. False

27. The common unit of power or rate of doing work is horsepower. This· is equal to 746 ft. lbs/min.

*a. True

b. False

28. Total dynamic head is the sum of the suction head and the discharge head minus the friction head

a. True

*b. False

29. Water power is the output power of the pump

*a. True

b. False

30. A positive displacement pump should be started up with the discharge valve closed in order to

avoid any problems with "air lock".

a. True

*b. False

31. Brake horse power is the input power to the motor

a. True

*b. False

32. Cost of electrical usage for pumps is based upon kilowatt per hour

a. True

*b. False

33. The common unit of power or rate of doing work is horsepower. This· is equal to 746 ft. lbs/min.

*a. True

344

Chapter 21. Pumping Efﬁciencies and Costs

b. False

34. Total dynamic head is the sum of the suction head and the discharge head minus the friction head

a. True

*b. False

35. Water power is the output power of the pump

*a. True

b. False

36. What is the vertical distance between the elevation of the free water surface at the suction and that

of the free water surface at the discharge of a pump called?

a. Discharge head.

b. Dynamic head.

c. Velocity head.

*d. Static head.

37. If a wasting pump has a ﬁxed pump rate of 250 GPM, and your calculation indicates you must

waste 126,000 gallons, what hourly cycle rate do you set the timer?

a. Turn pump on 21 minutes every day

b. Turn pump on 504 minutes every hour

c. Turn pump on 42 minutes every day

*d. Turn pump on 21 minutes every hour

38. A positive displacement pump is connected to a 25’ wide x 125’ long x 12’ side water depth aer-

obic digester. How long will it take to empty the contents of the digester if the pump rate is 225

gallons per minute?

a. 15.3 hours

b. 2.8 hours

*c. 20.8 hours

d. 15.6 hours

39. A centrifuge is fed sludge with a concentration of 3.4% solids. If the sludge feed rate is set at 50

gallons per minute, what is the centrifuge loading rate in pounds per hour?

a. 763 lbs/hour

*b. 850 lbs/hour

c. 735 lbs/hour

d. 960 lbs/hour

40. Calculate the surface loading rate for a treatment plant with 4 clariﬁers each with a 100 foot diame-

ter. The plant has an inﬂuent ﬂow of 35 MGD.

a. 279 gal/sq ft

b. 950 gal/sq ft

c. 4,459 gal/sq ft

*d. 1,115 gal/sq ft

41. Calculate the ﬂow velocity in feet/minute if 7.5 MGD of ﬂow passes through a channel that is 3’

wide x 4’ deep, and the depth of ﬂow is 15 inches.

*a. 186 ft/min

b. 58 ft/min

c. 202 ft/min

d. 46.5 ft/min

42. Determine the pounds per day of primary solids removed at a plant with a ﬂow rate of 1.5 MGD

21.5 Practice Problems

345

and the following data:

Clear Waters Fall 2013

Inﬂuent TSS = 250 mg/L

Primary Efﬂuent TSS = 150 mg/L,

Final Efﬂuent TSS = 12 mg/L

a. 1,101 lbs/day

*b. 1,251 lbs/day

c. 982 lbs/day

d. 2,977 lbs/day

43. A sewage pump is located above the wet well which is 8 feet deep and the pump is pumping to an

above ground clariﬁer with 12 feet depth of water. The pump manufacturer has given you the pump

characteristics curve which shows Total Dynamic Head vs. ﬂow rates. If the operating wet well

water depth is 6 feet, what is the total dynamic head in order to determine pumping rate from the

chart? Assume the top of the wet well and the bottom of the clariﬁer are at the same elevation.

a. 12 feet

b. 20 feet

*c. 14 feet

d. 10 feet

44. A sewage pump is located above the 8-foot diameter wet well which is 8 feet deep and the pump is

pumping to an above ground clariﬁer. The ﬂow meter on the pump is not operating and you want to

calculate the pumping rate by measuring the drop in wet well water level during when inﬂow to wet

well is minimal? If the drop in water level in one minute is 2 feet, what is the approximate pumping

rate in gallons per minute?

a. 250 GPM

b. 375 GPM

c. 500 GPM

*d. 750 GPM

22. Water Math

22.1 Fractions

•

A fraction is deﬁned as part of whole. If in a class there are 20 male students and 30 male students,

the fraction of male students is or .

• It is composed of three items: two numbers and a line.

•

The number on the top is called the numerator, the number on the bottom is called the denominator,

and the line in between them means to divide.

Numerator

Divide −→

Denominator

•

A proper fraction is a fraction that has no whole number part and its numerator is smaller than its

denominator. An improper fraction is a fraction that has a larger numerator than denominator and it

represents a number greater than one.

1 5 11

Proper Fraction Examples:

, ,

2 8 12

12 5

Improper Fraction Examples:

•

Any whole number can be expressed as a fraction by placing a "1" in the denominator. For exam-

ple:

2 is the same as and 45 is the same as

Only fractions with the same denominator can be added/subtracted, and only the numerators are

added/subtracted. For example:

and,

−

348

Chapter 22. Water Math

• A fraction combined with a whole number is called a mixed number. For example:

4 , 16 , 8 , 45 and, 12

These numbers are read, four and one eighth, sixteen and two thirds, eight and three fourths, forty-

ﬁve and one half, and twelve and seventeen thirty seconds.

Mized numbers

•

A fraction can be changed by multiplying the numerator and denominator by the same number.

This does not change the value of the fraction, only how it looks. For instance:

is the same as

which is

• Steps to convert

to a mixed number:

Step 1. How many times can 4 ﬁt into 17? 4 because 4×4=16. Thus, 4 becomes the whole number

part

Step 2. How much is left over in the numerator? 1 because 17−16 = 1. Thus, 1 becomes the numera-

tor of the fractional part

Step 3.

= 4

•

To turn a mixed number into an improper fraction, multiply the whole number part by the denomi-

nator and add the numerator. This becomes the new numerator over the original denominator.

Example: Converting 1.5 feet to fraction

1.5 ft = 1

1∗2+1 2+1

•

A mixed value - say a circumference is given in feet and fraction of feet (say 7 3/4), needs to be

converted to a fraction for calculation purposes.

22.2 Ratio

• Ratio is used for comparing the size of two or more quantities.

•

Say if there are 10 red cubes and 5 pink marbles in a bag, the ratio

and red cubes. It can also be represented by 5:10.

is the ratio of pink marbles

• 5 lbs of chemical in 10 gallons solution is a ratio. So is 30 miles per gallon.

• Unlike fractions, ratio does not compare things that have the same units.

22.3 Proportion

• Two quantities are said to be in proportion if one changes, the other changes in a speciﬁc way.

•

Two quantities are said to be directly proportional, if the increase in one will increase the other

value proportionally.

–

Thus, if two quantities x and y are directly proportional, its ratio will be a ﬁxed value. Thus

x₁

y₁

for x₁and y₁different values of x and y respectively will be related by the equation

22.3 Proportion

349

–

This relationship is useful for calculating unknown values in water treatment calculations as

in the following example:

Knowing 200 lbs of bleach is needed to disinfect 5 MG of water at a treatment plant, calcu-

late the lbs of bleach required to disinfect 3.2 MGD ﬂow.

200 pounds bleach

The ratio

or 40 lbs bleach per MG is a constant. Using this known

5MG

proportion the lbs of bleach is needed to disinfect 3.2 MG at this plant can be calculated as

follows by setting up the equation as:

40 pounds bleach

where X is the unknown lbs of bleach that is required to

3.2 MG

disinfect the 3.2 MG ﬂow.

3.2∗40

X can be calculated by cross multiplying the above equation: X =

= 128 lbs bleach

•

Two quantities are said to be inversely proportional if the increase in one will decrease the other

value proportionally.

–

Thus, if two quantities x and y are inversely proportional, its product x∗y will be a ﬁxed value

and different values of x and y respectively will be related by the equation x∗y = x₁∗y₁.

– Examples of inversely proportional relationship include:

Labor hours required to perform a certain task or time required to pump down a wetwell

depending on the size of the pump. An increase in assignment of labor hours will reduce

the time required to perform the task

Using a larger pump will reduce the time to pump down the wetwell.

In the Pounds formula:

ꢀ

ꢁ

lbs

day

lbs or

= Concentration

∗8.34∗volume(MG) or Flow(MGD)

for the same lbs or lbs/day, concentration varies inversely with volume or ﬂow. Thus, for

a certain pounds added, the concentration will go down if the ﬂow increases and vice

versa.

In the ﬂow equation, Q=V*A, for the same ﬂow (Q), velocity (V) and surface area (A)

are inversely related. If Q is remaining the same, an increase in surface area will reduce

the velocity and vice versa.

Additionally, for a ﬂow through a pipe as the surface area of the pipe is proportional to

the square of the diameter, the velocity in the pipe is inversely proportional to the square

of the diameter.

For a constant Q: V ∗A = V₁∗A₁or V ∗D²= V₁∗D₁²

–

Application of inversely proportional relationship in water related calculation can be demon-

stration with the following example:

If it takes 20 minutes to pump a wet well down with one pump pumping at 125 gpm, then

how long will it take if a 200 gpm pump is used?

As this is an inversely proportional relationship ( a larger pump will reduce the time re-

quired):

(20minutes∗125gpm) = (Xminutes∗200gpm)

where X is the unknown time to pump down the wetwell with the 200 gpm pump.

350

Chapter 22. Water Math

20∗125

Solving for X: X =

= 12.5 minutes

200

22.4 Decimals & Powers of Ten

•

A decimal is composed of two sets of numbers: the numbers to the left of the decimal are whole

numbers, and numbers to the right of the decimal are parts of whole numbers, a fraction of a num-

ber.

•

The term used to express the fraction component is dependent on the number of characters to the

right of the decimal.

– The ﬁrst character after the decimal point is tenths: 0.1 - tenths

– The second character is hundredths: 0.01 - hundredths

– The third character is thousandths: 0.001 - thousandths

• Powers of 10 notation enables us to work with these very large and small quantities efﬁciently.

•

In water math, the most common application of this concept is related to parts per million (ppm) or

parts per billion (ppb).

•

1 million - 1,000,000 can be represented as 10⁶. Likewise, 1 billion - 1,000, 000,000 can be repre-

sented as 10⁹

• The sequence of powers of ten can also be extended to negative powers.

• 1 part per million (1/1,000,000) can be written as 10⁻6

Name

one

Power

10⁰

Number

SI symbol SI preﬁx

ten

10¹

10²

10³

10⁶

10⁹

10⁻¹

10⁻²

da (D)

deca

hecto

kilo

hundred

thousand

million

billion

tenth

100

h (H)

1,000

k (K)

1,000,000

1,000,000,000

0.1

mega

giga

deci

hundredth

thousandth 10⁻³

0.01

centi

milli

micro

nano

0.001

millionth

billionth

10⁻⁶

10⁻⁹

0.000 001

0.000 000 001

22.5 Rounding and Signiﬁcant Digits

•

Signiﬁcant digits (also called Signiﬁcant Figures) are digits which give us useful information about

the accuracy of a measurement and are related to rounding.

•

This concept is used to determine the direction to round a number (answer). The basic idea is that

no answer can be more accurate than the least accurate piece of data used to calculate the answer.

• Signiﬁcant digits is the count of the numerals in a measured quantity (counting from the left) whose

values are considered as known exactly, plus one more whose value could be one more or one less.

• Rules for determining the number of signiﬁcant digits:

22.5 Rounding and Signiﬁcant Digits

351

1. All nonzero digits are signiﬁcant:

1.234 g has 4 signiﬁcant ﬁgures, and 1.2 g has 2 signiﬁcant ﬁgures.

2. Zeroes between nonzero digits are signiﬁcant: 1002 kg has 4 signiﬁcant ﬁgures, 3.07 mL has

3 signiﬁcant ﬁgures.

3. Zeroes to the left of the ﬁrst nonzero digits are not signiﬁcant; such zeroes merely indicate

the position of the decimal point: 0.001 ^◦C has only 1 signiﬁcant ﬁgure, 0.012 g has 2 signiﬁ-

cant ﬁgures.

4. Zeroes to the right of a decimal point in a number are signiﬁcant: 0.023 mL has 2 signiﬁcant

ﬁgures, 0.200 g has 3 signiﬁcant ﬁgures.

5. When a number ends in zeroes that are not to the right of a decimal point, the zeroes are not

necessarily signiﬁcant: 190 miles may be 2 or 3 signiﬁcant ﬁgures, 50,600 calories may be

3, 4, or 5 signiﬁcant ﬁgures. The potential ambiguity in the last rule can be avoided by the

use of standard exponential, or ”scientiﬁc,” notation. For example, depending on whether 3,

4, or 5 signiﬁcant ﬁgures is correct, we could write 50,600 calories as: 5.06∗10⁴calories (3

signiﬁcant ﬁgures) 5.060 ∗ 10⁴calories (4 signiﬁcant ﬁgures), or 5.0600 ∗ 10⁴) calories (5

signiﬁcant ﬁgures).

• Examples of signiﬁcant ﬁgures:

1000 has one signiﬁcant digit: only the 1 is interesting (only it tells us anything speciﬁc); we don’t know anything for sure about the hundreds,

tens, or units places; the zeroes may just be placeholders; they may have rounded something off to get this value.

1000.0 has ﬁve signiﬁcant digits: the ".0" tells us something interesting about the presumed accuracy of the measurement being made;

namely, that the measurement is accurate to the tenths place, but that there happen to be zero tenths.

0.00035 has two signiﬁcant digits: only the 3 and 5 tell us something; the other zeroes are placeholders, only providing information about

relative size.

0.000350 has three signiﬁcant digits: the last zero tells us that the measurement was made accurate to that last digit, which just happened to

have a value of zero.

1006 has four signiﬁcant digits: the 1 and 6 are interesting, and we have to count the zeroes, because they’re between the two interesting

numbers.

560 has two signiﬁcant digits: the last zero is just a placeholder.

560. : notice that "point" after the zero! This has three signiﬁcant digits, because the decimal point tells us that the measurement was made to

the nearest unit, so the zero is not just a placeholder.

560.0 has four signiﬁcant digits: the zero in the tenths place means that the measurement was made accurate to the tenths place, and that there

just happen to be zero tenths; the 5 and 6 give useful information, and the other zero is between signiﬁcant digits, and must therefore also be

counted.

• Addition and Subtraction

–

When you are adding or subtracting a bunch of numbers and need to be concerned with

signiﬁcant ﬁgures, ﬁrst add (or subtract) the numbers given in their entire format, and then

round the ﬁnal answer. When rounding the ﬁnal answer after adding or subtracting, the answer

must be written with the same significant ﬁgures as the least accurate decimal place given.

Example: 13.214 + 234.6 + 7.0350 + 6.38

13.214 + 234.6 + 7.0350 + 6.38 = 261.2290

234.6 is only accurate to the tenths place making it the least accurate number. My an-

swer must be rounded to the same place as the least accurate number:

261.2290 rounds to 261.2 (one decimal place)

• Multiplication and Division

–

When multiplying or dividing multiple numbers you would do these calculations as normal.

352

Chapter 22. Water Math

When the answer must be written in the appropriate signiﬁcant ﬁgure your answer must

round to the same number of significant ﬁgures as the least number of significant ﬁgures.

Example 1: Simplify, and round, to the appropriate number of signiﬁcant digits

16.235 x 0.217 x 5

Step 1. First off, 5 has only one signiﬁcant ﬁgure, thus the ﬁnal answer needs to be rounded to

one signiﬁcant digit

Step 2. 16.235 x 0.217 x 5 = 17.614975

Step 3. To round 17.614975 to one digit. I’ll start with the 1 in the tens place. Immediately to

its right is a 7, which is greater than 5, so 1 is rounded up to 2, and then replacing the 7

with a zero, and dropping the decimal point and everything after it.

Step 4. 17.614975 rounds to 20

Example 2: Simplify, and round, to the appropriate number of signiﬁcant ﬁgures

0.00435 x 4.6

Step 1. 4.6 has only 2 signiﬁcant ﬁgures, so the ﬁnal answer should be rounded to two signiﬁ-

cant ﬁgures.

Step 2. 0.00435 x 4.6 = 0.02001

Step 3. 0.02001 would round to 0.020, which has 2 signiﬁcant ﬁgures (0.020). The answer

cannot be 0.02, because that value would have only one signiﬁcant ﬁgure.

A number is rounded off by dropping one or more numbers from the right and adding zeroes, if

necessary, to maintain the decimal point.

•

• If the last ﬁgure dropped is 5 or more, increase the last retained ﬁgure by 1. If the last digit dropped

is less than 5, do not increase the last retained ﬁgure.

22.6 Averages

•

Also known as arithmetic mean, this value is arrived at by adding the quantities in a series and

dividing the total by the number in the series.

Example 1: Find the average of the following series of numbers: 12,8,6,21,4,5 , 9 , and 12.

Adding the numbers together we get 77.

There are 8 numbers in this set.

Divide 77 by 8.

= 9.6 is the average of the set

Example 2: Find the average of the set of daily turbidity data - 0.3,0.4,0.3,0.1,and 0.8

The total is 1.9.

There are 5 numbers in the set.

Therefore:

1.9

= 0.38, rounding off = 0.4

22.7 Working with Percent

• Percent expresses portions of the whole.

•

The whole is considered as 1 or 100% and a part of the whole can be expressed as a percent. Example:

If a tank is 1/2 full, we say that it contains 50% of the original solution.

• Percentage is written as a whole number with a % sign after it.

22.7 Working with Percent

353

• In a calculation, percent is expressed as a decimal.

• The decimal form of a percent value is obtained by dividing the percent by 100.

Example: 11% is expressed as the decimal 0.11, since 11% is equal to 11/100. This decimal is

obtained by dividing 11 by 100.

• To determine what percentage a part is of the whole, divide the part by the whole.

Example: There are 80 water meters to read, Jim has ﬁnished 24 of them. What percentage of the

meters have been read?

24÷80 = 0.30

The 0.30 is converted to percent by multiplying the answer by 100.

0.30×100 = 30%

Thus 30% of the 80 meters have been read.

•

To ﬁnd the percentage of a number, multiply the number by the decimal equivalent of the percentage

given in the problem.

Example:

What is 28% of 286?

Step 1. Change the 28% to a decimal equivalent:

28%÷100 = 0.28

Step 2. Multiply 286×0.28 = 80

Thus 28% of 286 is 80.

•

To increase a value by a percent, add the decimal equivalent of the percent to " 1 " and multiply it

times the number.

Example: A ﬁlter bed will expand 25% during backwash. If the ﬁlter bed is 36 inches deep, how

deep will it be during backwash?

Step 1. Change the percent to a decimal.

Step 2. Add the whole number 1 to this value.

Step 3. Multiply times the value.

25%÷100 = 0.25

1+0.25 = 1.25

36 in ×1.25 = 45 inches

22.7.1 Percentage Concentrations

Above concepts are used for chemicals such as ﬂuoride and hypochlorites - the strength of the product as

used is commonly expressed as a percentage.

Example 1: A chlorine solution was made to have a 4% concentration. It is often desirable to determine

this concentration in mg/L. This is relatively simple: the 4% is four percent of a million.

354

Chapter 22. Water Math

To ﬁnd the concentration in mg/L when it is expressed in percent, do the following:

1. Change the percent to a decimal.

4%÷100 = 0.04

3. Multiply times a million.

0.04×1,000,000 = 40,000mg/L

We get the million because a liter of water weighs 1,000,000mg.1mg in 1 liter is 1 part in a million parts (

ppm).1% = 10,000mg/L.

Example 2: How much 65% calcium hypochlorite is required to obtain 7 pounds of pure chlorine?

65% implies that in every lb of calcium hypochlorite has 65% lbs of available chlorine.

0.65 lbs available chlorine

lb of calcium hypochlorite

0.65 lbs available chlorine

Therefore,

or conversely

(

lb of calcium hypochlorite 7 lb of available chlorine

(

=⇒ lbs calcium hypchlorite required =

∗

(

0.65 lbs available chlorine

(

= 10.8 lbs of calcium hypochlorite with 65%available chlorine is required

22.8 Area & Volume

Example 1: The ﬂoor of a rectangular building is 20 feet long by 12 feet wide and the inside walls are 10

feet high. Find the total surface area of the inside walls of this building

Solution:

22.8 Area & Volume

355

Ceiling=W*L

Wall - L*H

Wall - W*H

Floor=W*L

Wall - W*H

Height=10’

Wall - L*H

Length=20’

Width=12’

2 Walls W*H + 2 Walls L*H= 2∗12∗10 ft²+2∗20∗10 ft²

= 240+400 = 640ft²

2 Walls W*H + 2 Walls L*H + Floor + Ceiling= 2∗12∗10 ft²+2∗20∗10 ft²+2∗12∗20 ft²

= 240+400+480 = 1,120ft²

Example 2: How many gallons of paint will be required to paint the inside walls of a 40 ft long x 65

ft wide x 20 ft high tank if the paint coverage is 150 sq. ft per gallon. Note: We are painting walls only.

Disregard the ﬂoor and roof areas.

Solution:

Wall - L*H

Wall - W*H

Height=20’

Wall - W*H

Wall - L*H

Length=40’

Width=65’

2 Walls W*H + 2 Walls L*H = 2∗65∗20 ft²+2∗40∗20 ft²= 2,600+1,600 = 4,200ft²

ft²

gal

4,200ft²

=⇒ @150

paint coverage →

= 28 gallons Example 3: What is the circumference of

ft²

150

gal

a 100 ft diameter circular sedimentation tank?

Solution:

Circumference = π ∗D = 3.14∗100ft = 314ft

Example 4: If the surface area of a clariﬁer is 5,025ft², what is its diameter?

Solution:

Sur face area = ∗D²=⇒ 5025( ft²) = 0.785∗D²( ft²)

√

5025

0.785

=⇒ D²=

=⇒ D = 6401.3 = 80ft

Example 5: How many gallons of water would 600 feet of 6-inch diameter pipe hold, approximately?

Solution:

Diameter=6"

Length=600’

Volume = D²∗

ꢀ

ꢁ

gallons

L = 0.785∗

∗600 ft³∗7.48

= 881 gallons

ft³

356

Chapter 22. Water Math

22.9 Flow and Velocity

• Flow Rate - Q (volume/time) = velocity (distance or length traveled /time) * surface area

• Velocity is the speed at which the water is ﬂowing. It is measured in units of length/time – ft./sec.

•

Velocity of water ﬂowing through can be calculated by dividing the ﬂow rate by area of the ﬂow

stream.

volume or cubic length

time

sur face area in the direction of flow−square length

flow rate(

)

length

time

Velocity

For a ﬂow in a channel:

For a ﬂow in a pipe:

Example 1: If a chemical is added in a pipe where water is ﬂowing at a velocity of 3.1 feet per second,

how many minutes would it take for the chemical to reach a point 7 miles away?

Note - we want the answer in minutes

1 sec 5280 ft

min

60sec

Min =

∗

∗7miles∗

= 199min

3.1 ft

mile

Example 2: Find the ﬂow in cfs in a 6 -inch line, if the velocity is 2 feet per second.

1. Determine the cross-sectional area of the line in square feet. Start by converting the diameter of the

pipe to inches.

The diameter is 6 inches: therefore, the radius is 3 inches. 3 inches is 3/12 of a foot or 0.25 feet.

2. Now ﬁnd the area in square feet.

A = π ×r²

ꢂ

A = π × 0.25ft²

A = π ×0.0625ft²

A = 0.196ft²

22.10 Concentration

357

A = 0.785×D²

A = 0.785×0.5²

A = 0.785×.05×.05

A = 0.196ft²

3. Now ﬁnd the ﬂow.

Q = V×A

Q = 2ft/sec×0.196ft²

Q = 0.3927cfs or 0.4cfs

22.10 Concentration

•

Concentration is typically expressed as mg/l which is the weight of the constituent (mg) in 1 liter of

water.

•

As 1 liter of water weighs 1 million mg, a concentration of 1 mg/l implies 1 mg of constituent per 1

million mg of water or one part per million (ppm). Thus, mg/l and ppm are synonymous.

• Sometimes the constituent concentration is expressed in terms of percentage.

Example: 12.5% chlorine concentration solution.

100% would mean 1,000,000 mg/l or 1,000,000 ppm

1,000,000

=⇒ 1% would be

mg/l = 10,000 mg/l or 10,000 ppm

100

=⇒ 12.5% chlorine concentration is 125,000 mg/l or 125,000 ppm.

1% concentration = 10,000 ppm or

0.1% concentration = 1,000 ppm or

0.01% concentration = 100 ppm or

10% concentration = 100,000 ppm or

5% concentration = 50,000 ppm or

12.5% concentration = 125,000 ppm or

22.11 Density

•

Density is deﬁned as the weight of a substance per a unit of its volume. For example, pounds per

cubic foot or pounds per gallon.

• Here are a few key facts about density:

–

Density is measured in units of lb/ft3, lb/gal, or mg/L. Density of water = 62.4 lb/ft3 = 8.34

lb/gal.

22.12 Speciﬁc Gravity

• Speciﬁc gravity is the ratio of the density of a substance (liquid or solid) to the density water.

•

It is the ratio of the weight of the substance of a certain volume to the weight of water of the same

volume.

358

Chapter 22. Water Math

•

Any substance with a density greater than that of water will have a speciﬁc gravity greater than 1.0.

Any substance with a density less than that of water will have a speciﬁc gravity less than 1.0.

• Speciﬁc gravity examples:

– Speciﬁc gravity of water = 1.0

– Speciﬁc gravity of concrete = 2.5 (depending on ingredients)

– Speciﬁc gravity of alum (liquid @ 60°F) = 1.33

– Speciﬁc gravity of hydrogen peroxide (35%) = 1.132

• Speciﬁc gravity is used in two ways:

1. To calculate the total weight of a % solution (either as a single gallon or a drum volume).

Total Weight = Drum Vol X SG X 8.34

2. To calculate the “active ingredient” weight of a single gallon or a drum.

Active Ingredient Weight within Drum = Drum Volume X SG X 8.34 X % solution as a

decimal. (i.e., Total Weight X % solution as a decimal)

NOTE: Both ways start with solving for the total weight (Drum Vol X SG X 8.34). When

solving for “active ingredient” weight, you have to then multiply by % solution as a decimal.

Example: What is the weight of 5 gallons of a 40% ferric chloride solution given its speciﬁc gravity of

1.43?

(8.34∗1.43) lbs/gal ∗5 gallons = 59.6 lbs

The weight of active ferric chloride in the drum will be 59.6*0.4=23.84 lbs (as ferric chloride is 40%

strength)

22.13 Detention Time

• Detention Time - The actual or theoretical (calculated) time required for water to ﬁll a tank at a

given ﬂow; pass through a tank at a given ﬂow; or remain in a settling basin, ﬂocculating basin,

rapid-mix chamber, or storage tank.

Tank/clari fier volume(cu. ft or gal)

Tank/clarifier detention time (hr) =

Influent flow (cu. ft or gal)/hr)

Rectangular tank/clariﬁer volume = width * length * depth of water

Circular tank/clariﬁer volume = 0.785 * Diameter²* depth of water

Typically volume is calculated in cu. ft and inﬂuent ﬂow is given in gallons. Use 7.48 gal/ft³con-

version factor to convert volume in cu. ft to gallons.

22.14 Unit Conversions

•

A conversion is a number that is used to multiply or divide into a measure in order to change the

units of the original measure.

In most instances, the conversion factor cannot be derived. It must be known. Therefore, tables

such as the one below are used to ﬁnd the common conversions.

22.14 Unit Conversions

359

Measure

Length

Area

Units

inches, ft, miles

ft², acres

Volume

Density

Flow

ft³, gallons, acres-ft.

weight per volume, lbs/ft³, lbs/gallon

ft³/min, MGD, acres-ft/day

Table 22.1: Common units in water calculations

Some Common Conversions

Weight

Linear Measurements

1 inch = 2.54 cm

1ft³of water = 62.4lbs

1gal = 8.34lbs

1foot = 30.5 cm

1lb = 453.6grams

1 kg = 1000 g = 2.2lbs

1% = 10,000mg/L

1pound = 16ozdrywt

1ft³= 62.4lbs

1 meter = 100 cm = 3.281feet = 39.4 inches 1

acre = 43,560ft²

1yard = 3feet

Volume

Pressure

1gal = 3.78 liters

1ft = 7.48 gal

1 L = 1000 mL

1gal = 16cups

Flow

1ft of head = 0.433psi

1psi = 2.31ft of head

1cfs = 448gpm

1gpm = 1440gpd

• Common conversions in water related calculations include the following:

– gpm to cfs

– Million gallons to acre feet

– Cubic feet to acre feet

– Cubic feet of water to gallons

– gpm to MGD

– psi to feet of head

• Steps for unit conversion:

Step 1: Make sure the original unit is for the same measurement as the converted (desired) unit. So

if the original unit is for area, say in ft²the converted unit should be another area unit such as

in²or acre but it cannot be gallons as gallon is a unit of volume.

Note: Calculating the weight of a certain volume of water involves the use of density which

is the mass per volume - value in units including lbs/gallon or lbs/ft³

360

Chapter 22. Water Math

Step 2: Write down the conversion formula as:

Quantity in converted unit = Quantity (Original Unit)∗Conversion Factor

Conversion unit

(

Original unit

(

•

Note: If you wish to convert cubic feet of water to pounds, you have to use its density which is the

known mass per unit volume.

8.34 lbs

gallon

62.4 lbs

ft³

mass

ꢁ

mass o f water = Volume∗Density(

)

ꢁ

Volume

ꢁ

Example Problems:

1. Convert 1000 ft³to cu. yards

cu.yards

1000 ft³∗

= 37cu.yards

27 ft³

2. Convert 10 gallons/min to ft³/hr

Note: This involves use of two conversion factors - one for converting gallons to cubic feet and

another for converting minute to gallons.

ꢁ

10gallons

ft³

60min 80.2ft³

ꢁ

∗

ꢁ

min

7.48gallons

ꢁ

3. Convert 100,000 ft³to acre-ft.

100,000ft³∗

4. Convert 8 ft³of water to pounds.

acre− ft

= 2.3acre− ft

ꢁ

43,560 ft − ft

ꢁ

Here the conversion is from a volume (ft³) to a weight (lbs). It involves use of a standard correla-

tion of the volume of water to its weight - its density.

lbs

ft³

Weight o f water in lbs = 8ft³∗62.4(

) = 499.2 lbs

22.15 Pounds Formula

Pounds formula is used for:

•

Calculating the quantity in pounds of a particular wastewater constituent entering or leaving a

wastewater treatment process

• Calculating the pounds of chemicals to be added

So if the concentration of a particular constituent (in mg/liter) and the volume or ﬂow of wastewater is

given, one can calculate the amount of that constituent in pounds using the following – Pounds Formula:

lbs

day

lbs or

= concentration( )∗8.34∗volume(MG) or flow(

(MGD)

Figure 22.1: Pounds formula "nomograph"

There are three variables – (lbs, concentration and volume) and one constant (8.34) in the pounds for-

22.16 Process Removal Efﬁciency Calculations

361

mula. Knowing any of the two variables in the formula, one can calculate the third (unknown) variable by

rearranging the equation.

22.15.1 Example Problems

1. If the inﬂuent wastewater ﬂow is 5 MGD and the BOD concentration is 240 mg/l what is the daily

BOD loading in lbs/day?

Solution:

lbs BOD

day

10,000lbs

= 5MGD∗240mg/l ∗8.34 =

day

2. Calculate the lbs of solids in the primary sludge if the sludge ﬂow is 7500 gallons and the solids

concentration is 4.5%.

Solution

Applying lbs formula:

7500 MG

1,000,000

lbs solids =

∗4.5∗10,000∗8.34 = 2,815 lbs solids

Note:

1) 7500 gallons was converted to MG by dividing by 1,000,000

1MG

1,000,000 gallons

7500 gallons∗

2) 4.5% was converted to mg/l by multiplying by 10,000 as 1%=10,000mg/l

3. An operator dissolves 1,200 lbs of a chemical in 12,000 gallons of water, what is the resultant

concentration in mg/l, of the chemical solution?

Solution:

lbs

Concentration

Volume MG ∗ 8.34

1,200

11,990 mg

Concentration

or 1.2% solution

0.012 ∗ 8.34

Note:

1) 12,000 gallons was converted to MG by dividing by 1,000,000

1MG

1,000,000 gallons

12,000 gallons∗

22.16 Process Removal Efﬁciency Calculations

• Process removal rate or removal efﬁciency is the percentage of the inlet concentration removed.

•

It is used for quantifying the pollutant removal during wastewater treatment and is established

based upon the amount of a particular wastewater constituent entering and leaving a treatment

process.

Pollutant In−Pollutant Out

• Process Removal Rate (%) =

∗100

Pollutant In

•

If 10 units of a pollutant are entering a process and 8 units of pollutant are leaving (process re-

moves 2 units), then the process removal rate for that pollutant is (10-8)/10*100=20%. In this

example the process is 20% efﬁcient in removing that particular pollutant.

•

The amount of pollutant can be measured in terms of concentration (mg/l) or in terms of mass

loading (lbs). The pounds formula is used for calculating the mass loadings.

The above example is for calculating the removal efﬁciency using the inlet and outlet concentrations or

mass loading.

The methods below can be used for calculating either the inlet or outlet pollutant concentrations, if the

removal efﬁciency and the corresponding inlet or outlet concentrations are given.

Case 1: Calculating outlet conc. (X) given the inlet conc. and removal efﬁciency (RE%):

362

Chapter 22. Water Math

X mg/l (Unknown)

A mg/l (Given)

100 mg/l

Process

Out

(100−RE%) mg/l

Removal E f ficiency = RE% (Given)

Using the fact that if the inlet concentration was 100 mg/l, the outlet concentration would be 100 minus

the removal efﬁciency.

Out

X mg/l

A mg/l

100−RE%

Setup the equation as:

100

Calculate X using cross multiplication - if

=⇒ A = B∗

100−RE%

X mg/l = A mg/l ∗

100

Case 2: Calculating inlet conc. (X) given the outlet conc. and removal efﬁciency (RE%):

X mg/l (Unknown)

100 mg/l

A mg/l (Given)

Process

Out

(100−RE%) mg/l

Removal E f ficiency = RE% (Given)

Using the fact that if the inlet concentration was 100 mg/l, the outlet concentration would be 100 minus

the removal efﬁciency.

Out

X mg/l

A mg/l

100

100−RE%

Setup the equation as:

Calculate X using cross multiplication - if

=⇒ A = B∗

100

X mg/l = A mg/l ∗

100−RE%

Example Problems:

1. What is the % removal efﬁciency if the inﬂuent concentration is 10 mg/L and the efﬂuent concentra-

tion is 2.5 mg/L?

In−Out

10−2.5

Removal Rate(%) =

∗100 =⇒

∗100 = 75%

2. Calculate the outlet concentration if the inlet concentration is 80 mg/l and the process removal

efﬁciency is 60%

Solution:

80mg/l

Xmg/l

Process

Out

40mg/l

100mg/l

Removal E f ficiency = 60%

Actual Outlet(X) 100−60

Out

100

Actual Outlet(X)

=⇒

= 0.4

=⇒ Actual Outlet(X) = 0.4∗80 = 32mg/l

22.17 Pounds Formula

363

3. Calculate the inlet concentration if the outlet concentration is 80 mg/l and the process removal

efﬁciency is 60%

Xmg/l

80mg/l

40mg/l

Process

Out

100mg/l

Removal E f ficiency = 60%

Out

Actual inlet (X)

100

100−60

Actual inlet (X)

=⇒

= 2.5

Rearranging the equation: Actual inlet(X) = 2.5∗80 = 200mg/l

22.17 Pounds Formula

• Pounds formula:

ꢀ

ꢁ

lbs

day

lbs or

= Concentration

∗8.34∗volume(MG) or Flow(MGD)

• So if the concentration of a particular constituent (in mg/liter) and the volume or ﬂow of wastewater

is given, one can calculate the amount of that constituent or using this formula.

Important notes:

1. The unit of the constituent loading rate will be in lbs per the unit of time the ﬂow is expressed

in. So if the ﬂow is in MG per day the calculated loading rate will be in lbs/day. Likewise if the

ﬂow value used is in MG per minute, the calculated loading rate will be in lbs/min.

2. If volume is used, the calculated value will be the mass of the constituent in that volume. If ﬂow

is used, the calculated value will be the mass of the constituent in that ﬂow.

3. For the Pound Formula to work, the volume or ﬂow needs to be expressed in MG. Volume or

ﬂows in other units - gallons, ft³etc. needs to be converted to MG.

•

The formula assumes that all of the material found in water (TSS, BOD, MLSS, Chlorine, etc.)

weighs the same as water, that is, 8.34 pounds per gallon.

In the Pounds Formula, there are three variables – lbs, concentration and volume, and one constant

- 8.34. Knowing any of the two variables in the formula, one can calculate the third (unknown)

variable by rearranging the equation.

•

Davidson Pie provides a pictorial reference for calculating any unknown variable. If for example, if

Concentration is unknown, it can be calculated as follows:

lbs

day

lbs or

ꢀ

ꢁ

Concentration

8.34∗Volume(MG) or Flow(MGD)

• Likewise, if Volume (or Flow) is the unknown variable. it can be calculated as:

lbs

lbs or

day

ꢀ

ꢁ

Volume(MG) or Flow(MGD) =

Concentration

∗ 8.34

364

Chapter 22. Water Math

lbs or lbs/day

Concentration

mg/l

Volume(MG)

Flow(MGD)

8.34

Multiply

Figure 22.2: Davidson Pie

• Pounds formula is used for:

–

Calculating the quantity in pounds of a particular wastewater constituent entering or leaving a

wastewater treatment process

– Calculating the pounds of chemicals to be added

Example 1: If a 5 MGD ﬂow is to be dosed with 25 mg/l of a certain chemical, calculate the lbs/day that

chemical required.

Solution

Applying lbs formula:

lbs

day

lbs

= 5MGD∗250

∗8.34 = 1,042

day

Example 2: Calculate the lbs of chemical in 7,500 gallons of 4.5% active solution of that chemical.

Solution

Applying lbs formula:

7500

lbschemical =

MG∗4.5∗10,000∗8.34 = 2,815 lbs chemical

1,000,000

Note:

1) 7500 gallons was converted to MG by dividing by 1,000,000

1MG

1,000,000 gallon

7500 gallons∗

2) 4.5% was converted to mg/l by multiplying by 10,000 as 1%=10,000mg/l

22.18 Chemicals Related Math Problems

22.18.1 Chemical Dosing

• Use lbs formula to calculate the lbs of chemicals required

•

Using the calculated lbs chemical required value, calculate the amount of that chemical at the

concentration available

So for example, if asked how much many gallons per day of bleach solution (SG 1.2)containing 12.5%

available chlorine is required to disinfect a 10 MGD ﬂow of water given the required chlorine dosage of 7

mg/l.

22.18 Chemicals Related Math Problems

365

1. calculate the lbs of chlorine required using the lbs formula:

=10MGD ∗ 7

∗ 8.34 = 583.8 lbs chlorine per day

2. calculate the gallons of bleach which will provide the 583.8 lbs chlorine

Applying the lbs formula - note that 8.34 * SG will give the actual lbs/gal of bleach. If SG is not

provided, use only 8.34 lbs per gallon:

lbs bleach

day

gal

day

lbs bleach lbs chlorine

∗ 0.0125

gal lb bleach

583.8

= x

∗ 8.34∗1.2

gal

day

583.8

8.34∗1.2∗0.125

gal

day

=⇒ x

= 467

The above problem can be solved directly using the formula below given in the SWRCB Water

Treatment Exam Formula Sheet.

(MGD)∗(ppm or mg/l)∗8.34 lbs/gal

% purity∗Chemical Wt. (lbs/gal)

10∗7∗8.34

gal

day

GPD =

= 467

0.125∗(1.2∗8.34)

22.18.2 Chemical Dilution

•

Chemicals obtained in bulk are typically delivered in higher concentrations to reduce transportation

costs and need dilution to ensure proper mixing and dosage control.

• Thus, for dilution calculations, if:

C₁and V₁is the concentration and volume respectively of the concentrated chemical used for the

dilution, and

C₂and V₂is the concentration and volume of the product after dilution with water

As, the mass of the target chemical in the volume of the concentrated product used for dilution will

remain the same in the ﬁnal diluted product:

C₁* V₁= C₂* V₂.

• Thus, knowing C₁, C₂and V₂, we can calculate V₁as:

C₂∗V₂

V₁=

C₁

Example Problem:

How much initial volume of a 4% polymer solution is needed to make 3500 gallons of polymer at 0.25%

concentration.

Solution:

C_.25%∗V_.25%

0.25 ∗ 3500

V_4%

= 219gal

C_4%

So take 219 gallons of the 4% polymer and dilute to 3500 gallons to give a 0.25% polymer solution.

366

Chapter 22. Water Math

22.19 Preliminary Treatment

22.20 Preliminary Treatment Math Problems

Preliminary Treatment math problems relate to the following:

22.20.1 Channel Velocity and Flow Rate

Flow Rate - Q (volume/time) = velocity (distance or length traveled /time) * surface area

Velocity is the speed at which the water is ﬂowing. It is measured in units of length/time – ft./sec.

Velocity of water ﬂowing through can be calculated by dividing the ﬂow rate by area of the ﬂow stream.

volume or cubic length

time

sur face area in the direction of flow−square length

flow rate(

)

length

time

Velocity

For a ﬂow in a channel:

Example Problems:

1. Calculate the velocity of a 14 MGD ﬂow in a 6 ft wide channel with a water depth of two feet.

Flow(Q) = Velocity(V)∗Area(A)

MG 10⁶gal

ft³

day

sec

ft³

sec

ft³

sec

=⇒ 14

∗

= V

∗6 ft ∗2 ft =⇒ 21.7

= 12V

day

21.7

7.48gal 24∗60∗60

=⇒ V

= 1.8

sec

2. Calculate the ﬂow, in gpd, that would pass through a grit chamber 2 feet wide, at a depth of 6

inches, with a velocity of 1 ft /sec

Solution:

Q = V ∗A

ft³

Q = 1 ∗(2∗0.5)ft²= 1

ft³(1440∗60)s

gal

ft³

gal

day

Q = 1

∗

∗7.48

= 646,272

day

3. A wastewater channel is 3.25 feet wide and is conveying a wastewater ﬂow of 3.5 MGD. The

wastewater ﬂow is 8 inches deep. Calculate the velocity of this ﬂow.

22.21 Primary Treatment

367

Solution:

Q = V ∗A =⇒ V =

day

ꢀ

MG 1000000gal

ꢀ

3.5

∗

ꢀ

7.48gal

day

(1440∗60)s

ꢀ

=⇒ V

= 2.2

(3.25∗0.75) ft²

4. A plastic ﬂoat is dropped into a wastewater channel and is found to travel 10 feet in 4.2 seconds.

The channel is 2.4 feet wide and is ﬂowing 1.8 feet deep. Calculate the ﬂow rate of this wastewater

in cubic feet per second.

Solution:

Q = V ∗A

ꢀ

ꢁ

ft³

10 ft

4.2s

ft³

=⇒ Q

∗(2.4∗1.8) ft²= 10.3

22.20.2 Grit Removal Rates

Typical grit removal ranges from 0.5 to 30 ft³/MG

Example Problems:

1. At a wastewater treatment plant which receives a ﬂow rate of 650,000 gallons per day, a total of 50

cubic feet of grit was removed for the month. Calculate the rate of grit removal assuming 30 days

in a month.

Solution:

ꢁ

ft³

ꢁmonth

gal

ft³

ꢁ

day

ꢀ

GritRemoval

= 50

∗

∗1,000,000

= 2.6

ꢁ

ꢀ

ꢁmonth 30days

650,000gal

ꢀ

22.21 Primary Treatment

22.21.1 Hydraulic or Surface Loading Rate

The hydraulic or surface loading rate measures how rapidly wastewater moves through the primary clar-

iﬁer. It is measured in terms of the number of gallons ﬂowing each day through one square foot surface

area of the clariﬁer.

ꢀ

ꢁ

gpd

ft²

Clari fier in fluent flow(gpd)

Clari fier sur face area( ft²)

Clarifier hydraulic loading

Rectangular clariﬁer surface area = width * length

Circular clariﬁer surface area = 0.785 * Diameter²

22.21.2 Detention Time

Detention time is the length of time that wastewater stays in the settling tank is called the detention time.

It is also the time it takes for a unit volume of wastewater to pass entirely through a primary clariﬁer

368

Chapter 22. Water Math

Clari fier volume(cu. ft or gal)

Clarifier detention time (hr) =

Influent flow (cu. ft or gal)/hr)

Rectangular clariﬁer volume = width * length * depth of water

Circular clariﬁer volume = 0.785 * Diameter²* depth of water

Typically volume is calculated in cu. ft and inﬂuent ﬂow is given in gallons. Use 7.48 gal/ft³conversion

factor to convert volume in cu. ft to gallons.

22.21.3 Weir Overﬂow Rate

The weirs at the end of the primary clariﬁer allow for the even distribution of the the outlet ﬂow across

the entire length of the weir. An adequate length of weir is needed to ensure smooth and even ﬂow of

wastewater over the weirs. Weir overﬂow rate measures the number of gallons of wastewater per day

ﬂowing over one foot of weir.

ꢀ

ꢁ

ꢀ

ꢁ

gpd

Clari fier in fluent flow(gpd)

Total e f fluent weir length ( ft)

Weir over flow rate

Circular clariﬁer weir length = 3.14 * Diameter

Example problem for (a), (b) and (c) above:

A circular clariﬁer receives a ﬂow of 11 MGD. If the clariﬁer is 90 ft. in diameter and is 12 ft. deep, what

is: a) the hydraulic/surface loading rate, b) clariﬁer detention time in hours, and c) weir overﬂow rate?

a) Hydraulic/surface loading rate:

11MG 10⁶gal

∗

ꢀ

ꢁ

gpd

ft²

day

0.785∗90²ft²

Clarifier hydraulic loading

= 1,730gpd/ft²

b) Clariﬁer detention time:

Clarifier volume(cu. ft or gal)

In fluent flow (cu. ft or gal)/hr)

Clarifier detention time (hr) =

(0.785∗90²∗12)ft³

Clarifier detention time (hr) =

= 1.2hrs

ft³

day

ꢀ

11MG 10 gal

ꢀ

∗

ꢀ

24hrs

7.48gal

day

ꢀ

c) Weir overﬂow rate:

11MG 10⁶gal

∗

ꢀ

ꢁ

gpd

day

3.14∗90 ft

Weir over flow rate

= 38,924gpd/ ft

22.21.4 Removal Efﬁciency

Primary sedimentation removes suspended wastewater solids which includes BOD. The efﬁciency of

the primary is established as the percentage of the amount of parameter removed. The parameter may

quantiﬁed as mass (lbs) or as concentration (mg/l).

Parameter In−Parameter Out

Removal e f ficiency(%) =

∗100

Parameter In

For TSS removal:

22.21 Primary Treatment

369

TSS_In(mg/l)−TSS_Out(mg/l)

TSS_In(mg/l)

TSS Removal e f ficiency(%) =

For BOD removal:

BOD Removal e f ficiency(%) =

22.21.5 Solids Removal

∗100

BOD_In(mg/l)−BOD_Out(mg/l)

BOD_In(mg/l)

∗100

Type 1 Problems: These involve calculating lbs of solids removed given any two of the following TSS

parameters - inlet concentration, outlet concentration and removal efﬁciency.

a. If the inlet and outlet concentrations are given, calculate the mg/l of TSS removed using:

TSS_removed= TSS_in(mg/l)−TSS_out(mg/l)

Then knowing the ﬂow, use the lbs formula to calculate the lbs solids removed.

b. If either inlet or outlet concentration is given along with the clariﬁer removal efﬁciency, using the

removal efﬁciency calculate the unknown outlet concentration (if only the inlet is given) or the inlet

concentration (if only the outlet is given)

i) If inlet and removal efﬁciency is given, calculate the outlet by subtracting the product of inlet and

removal efﬁciency from the inlet.

TSS_out= TSS_in−(TSS_in∗%Removal)

Example if the removal efﬁciency is 60% and the inlet concentration is 300mg/l:

TSS_out= 300−300∗0.6 = 120mg/l

ii) If outlet and removal efﬁciency is given, calculate the inlet concentration by dividing the outlet by

(1-removal efﬁciency).

TSS_out

TSS_in=

1−%Removal

Example if the removal efﬁciency is 60% and the outlet concentration is 120mg/l:

120

1−0.6

TSS_in=

= 300mg/l

TSS_in−TSS_out

Note: You may derive the above formulas by algebraically manipulating: %Removal =

TSS_in

Example Problem:

How many lbs of solids are removed daily by a primary clariﬁer treating a 6 MGD ﬂow if the average

inﬂuent TSS concentration is 300 mg/l and the clariﬁer TSS removal efﬁciency is 67%.

TSS_out= (300mg/l −300∗0.67) = 99mg/l

lbs solids removed = (300−99)mg/l ∗8.34∗6MGD = 10,058 lbs solids removed per day

Type 2 Problems: These involve calculating the amount of sludge pumping given the solids removed. The

solids removed from the primary clariﬁer is sludge with a typical solids concentration of about 3% to 5%.

370

Chapter 22. Water Math

Given the amount of total solids removed and given the sludge concentration, the volume of sludge pump-

ing can be calculated as follows:

ft³sludge pumped

day

lbs solids (removed)

1 lb sludge

gal sludge

ft³sludge

∗

∗ ∗

(%) lbs solids 8.34lb sludge 7.48 gal

day

So for the solids removed in the above example, if the primary sludge has 5% solids, the required sludge

pumping can be calculated as:

(

ꢁ

(

ꢁ

(

ft³sludge 10,058 lbs solids

1 lb sludge

ga(l sludge

ft³sludge

day

(

ꢁ

(

ꢁ

∗ ∗ ∗

ꢁ

(

0.05 lbs solids 8.34lb sludge

(

ꢁ

= 3,224

(

ꢁ

(

ꢀ

day

7.48 gal

ꢀ

22.22 Trickling Filter Calculations

Trickling ﬁlter problems involve calculation of the following:

22.22.1 Hydraulic or surface loading

• Hydraulic or surface loading is expressed as gpd/ ft²

• The gpd is the total ﬂow (Q_T)to the ﬁlter - primary inﬂuent ﬂow + recirculated ﬂow(Q_T= Q_I+ Q_R)

Example Problem:

The total inﬂuent ﬂow (including recirculation) to a trickling ﬁlter is 1.89 MGD. If the trickling ﬁlter is 80

ft in diameter, what is the hydraulic loading in gpd/sq ft on the trickling ﬁlter?

Solution:

gpd

ft²

(1.89∗10⁶)gpd

(0.785∗80²)ft²

gpd

ft²

Hydraulic loading

= 376

22.22.2 BOD and TSS Removal

BOD and removal is based upon the TF inﬂuent and efﬂuent concentrations.

In_conc−Out_conc

%Removal =

∗100

In_conc

Example Problem:

The suspended solids concentration entering a trickling ﬁlter is 236 mg/l. If the suspended solids con-

centration of the trickling ﬁlter efﬂuent is 33 mg/l, what is the suspended solids removal efﬁciency of the

trickling ﬁlter?

Solution:

236mg/l −33mg/l

%Removal =

∗100 = 86%

236mg/l

22.22.3 Organic Loading

• Organic loading to a trickling ﬁlter is typically expressed as lbs BOD/(day-1000 cu ft).

• The lbs/day BOD value is the BOD loading from the primary efﬂuent.

• The 1000 ct. ft is the volume of the media.

• The media volume is calculated by multiplying the TF surface area by the media height.

•

As the dimensions are typically given in ft., calculate the volume in ft³and then divide the calcu-

lated volume by 1000 to give the volume in units of 1000ft³

22.23 Stabilization Pond Calculations

371

Example Problem:

A trickling ﬁlter, 70 ft in diameter with a media depth of 6 ft, receives a ﬂow of 0.78 MGD. If the BOD

concentration of the primary efﬂuent is 167 mg/L, what is the organic loading on the trickling ﬁlter in lbs

BOD/day/1000 cu ft?

lbs BOD

lbs BOD feed to TF per day

Solution: Organic loading :

day−1000ft³

volume in 1000ft³

(0.78∗167∗8.34)lbs BOD

47lbs BOD

day

1000 ft³

(0.785∗70²∗6) ft³∗

1000 ft³

day−1000 ft³

22.22.4 Recirculation Ratio

Recirculated Flow(Q_R)

In fluent Flow(Q_I)

Recirculation Ratio(R_R) =

Total Flow(Q_T)−Influent Flow(Q_I)

In fluent Flow(Q_I)

Recirculation Ratio(R_R) =

Total Flow(Q_T) = In fluent Flow(Q_I)∗(Recirculation Ratio(R_R)+1) Make sure Q_R

are the same in a given problem

, Q_Tand Q_Iunits

Example Problems:

1. The inﬂuent to the trickling ﬁlter is 1.61 MGD. If the recirculated ﬂow is 2.27 MGD, what is the

recirculation ratio?

Q_R

Q_I

2.27

1.61

Solution: R_R=

= 1.4

2. A trickling ﬁlter has a total ﬂow of 32 MGD. If the recirculation ratio is 0.8, what is the primary

efﬂuent ﬂow to the TF?

Solution:

Total Flow(Q_T) = In fluent Flow(Q_I)∗(Recirculation Ratio(R_R)+1)

1.8

=⇒ 32MGD = Q_I∗(0.8+1) =⇒ Q_I=

= 17.8MGD

22.23 Stabilization Pond Calculations

22.23.1 Pond Area

Formula: Pond Area = Width∗Length

Pond Volume

Pond Depth

also, Pond Area =

Example Problem:

A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in acres?

Solution:

acre

43,560ft²

(260∗80) ft²∗

= 0.48acre

22.23.2 Solids Loading Rate

lbs TSS

day

Formula: Pond TSS loading rate =

Example Problem:

372

Chapter 22. Water Math

The inﬂuent ﬂow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L in

the raw wastewater. How many lbs of suspended solids are sent to the pond daily?

Solution:

lbs TSS

day

gal 24hrs

lbs TSS

∗8.34 = 284

day

= 10,000

∗

∗ ∗142

day 1,000,000gal

22.23.3 Organic Loading Rate

Formula: Pond organic loading rate =

Example Problem:

lbs BOD/day

Area(acre)

The ﬂow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond inﬂuent is 170mg/L,

what is the organic loading to this pond in lbs BOD/day/acre?

Solution:

lbs BOD per day (7.2MGD ∗ 170mg/l ∗ 8.34) 43,560ft²

4,624lbs BOD

day−acre

Organic loading =

∗

area (acres)

0.785∗350²ft²

acre

22.23.4 Detention time

Volume

Flow

Formula: Pond detention time =

Example Problem:

A 40 acre wastewater treatment pond receives a ﬂow of 0.6 MGD. If the pond is operated at a depth of 4ft.

What is the detention time of this pond?

Solution:

Volume

Flow

(40∗4)acre− ft

ft³

Pond detention time =

= 87 days

gal

acre− ft

0.6∗10⁶

∗

day 7.48gal 43,560ft³

22.23.5 Hydraulic Loading Rate

day

Flow

Area

Formula:Pond hydraulic loading rate

day

Pond depth (in)

also, Pond hydraulic loading rate

Volume

Flow

Pond detention time

The second formula above is because:

Flow

Hydraulic Loading (HL) =

Area

Vol

Flow

Vol

Detention time (DT) =

=⇒ Flow =

Substituting for ﬂow in the HL formula above:

Vol

Area

Vol

Area∗DT

Pond Depth

Vol

Area

HL =

=⇒ HL =

= Pond Depth

Example Problems:

22.24 Activated Sludge Calculations

373

1. Find hydraulic loading in inches/day for a pond given the following:

• Pond depth = 12ft.

• Pond volume = 1,400,000ft3

• Pond ﬂow = 1,000,000gal/day

Solution:

day

Flow

Area

Pond hydraulic loading rate

ft³

gal

1,000,000

∗

∗12 = 13.8

day 7.48gal

1,400,000 ft³

12 ft

=⇒

day

Note: The area of the pond was found by dividing the volume (1,000,000 ft³) by the pond depth (12ft)

2. Find pond hydraulic loading in inches/day when the depth of the pond is 6 ft. and the detention

time is 30 days.

Solution:

Pond depth (in)

Volume

Pond detention time

Flow

Pond hydraulic loading rate =

6∗12 inches

30 days

2.4in

day

=⇒

22.24 Activated Sludge Calculations

22.24.1 Mean Cell Residence Time

The MCRT is calculated as:

MCRT(days) =

Total MLSS lbs in the aeration system (aeration tank + clari fier)

Total amount in lbs/day o f suspended solids leaving the system (E f fluent SS+WAS solids)

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

E f fluent suspended solids (lbs/day)+ WAS SS (lbs/day)

Key Points for Solving MCRT Problems

1. MLSS quantiﬁcation:

•

Pounds formula is used to calculate lbs MLSS using: i) aeration tank and the clariﬁer vol-

umes, and ii) the given MLSS concentration.

The MLSS concentrations for the aeration tank and the clariﬁer are the same. So the given

MLSS concentration applies to both - the aeration tank and the clariﬁer

• Make sure it is the MLSS concentration that you are using not the MLVSS concentration.

•

If MLSS concentration is not given but instead MLVSS concentration is given, you will need

to ﬁnd the MLSS concentration by dividing the MLVSS conc. by the mixed liquor volatile

solids, as MLVSS(conc.) = MLSS * % volatile solids

2. Suspended solids quantiﬁcation:

(a) Efﬂuent suspended solids

• Efﬂuent suspended solids can be quantiﬁed using the pounds formula - using the efﬂuent

ﬂow (in MGD) and the efﬂuent suspended solids concentration.

374

Chapter 22. Water Math

(b) WAS suspended solids

•

Use pounds formula given the WAS ﬂow (make sure it is in MG) and the WAS SS

concentration

•

Note that the WAS and RAS streams have the same SS concentration. If WAS SS con-

centration is not speciﬁed, use the RAS SS concentration

Example Problems:

1. In an conventional activated sludge plant the aeration tank contains 6000 lbs of MLSS and the ﬁnal

clariﬁer contains 2300 lbs of MLSS. 1450 lbs of solids are wasted each day and 90 lbs/day of solids

leave in the ﬁnal efﬂuent. Calculate the MCRT for this plant.

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

6000lbs + 2300lbs

90lbs/day + 1450lbs/day

MCRT(days) =

= 5.4 = 5days

2. A activated sludge plant treats an average inﬂuent ﬂow of 4 MGD. The plant has two aeration

tanks – 0.45 MG volume each and two ﬁnal clariﬁers – 0.2 MG volume each, and a mixed liquor

suspended solids concentration averages 1800 mg/l. The efﬂuent suspended solids concentration

averages 18 mg/L. The WAS ﬂow is 100,000 gallons per day has a SS concentration of 6100 mg/L.

Calculate the MCRT

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 2∗0.45∗1800∗8.34 = 13511lbs

MLSS in clarifier (lbs) = 2∗0.2∗1800∗8.34 = 6005lbs

SS e f fluent (lbs/day) = 4MGD∗18mg/l ∗8.34 = 600lbs/day

100000

1000000

SS WAS (lbs/day) =

MGD∗4800mg/l ∗8.34 = 4003lbs/day

13511+6005

Plugging in the values calculated above: MCRT(days) =

= 4.2 = 4days

600+4003

22.24.2 F:M (Food to Microorganism Ratio

•

This parameter ratios the food – the mass of primary efﬂuent BOD entering the aeration basin to

the mass of the microorganisms - MLVSS, in the aeration basin.

•

Only the mass of the microorganisms (MLVSS) in the aeration basin is used – the mass of

microorganisms in the secondary clariﬁer is not considered

• Common ranges for F/M for a conventional activated sludge plant are from 0.15 to 0.5.

• The optimum F/M varies from plant to plant and can be determined by trial and error.

•

The F:M may be used to determine the concentration of mixed liquor suspended solids to be main-

tained in the aeration tank.

Generally, low F/M ratios should be carried during the colder months as the microorganism activity

(metabolism) is lower.

F:M and MCRT are inversely related: that is a long MCRT means a low F:M and a short MCRT

means a high F:M

amount o f food coming in

amount o f microorganisms present

F:M=

22.24 Activated Sludge Calculations

375

(lbs/day) primary e f fluent BOD entering the aeration tank

(lbs) MLVSS in the aeration tank

Key Points for Solving F:M Problem

1. Quantifying F:

• use the pounds formula to calculate the lbs/day of BOD in the primary efﬂuent.

lbs/day BOD = Primary eff. ﬂow (MGD)* Primary eff. BOD concentration (mg/l) * 8.34

2. Quantifying M:

•

The concentration of the microorganisms is assumed to be the same as the MLVSS concentra-

tion

Only the mass of the microorganisms (MLVSS) in the aeration basin is used – the mass

of microorganisms in the secondary clariﬁer is not considered

lbs MLVSS may be calculated using pounds formula using the volume of the aeration tank

(in MG) and the MLVSS concentration

If the MLVSS concentration is not given, it can be calculated from the MLSS and MLSS %

volatile matter (solids) concentration

MLVSS = MLSS * % MLSS volatile solids

Example Problem:

i. A conventional activated sludge plant receives an average ﬂow of 5.5 MGD. The inﬂuent

BOD to the plant averages 230mg/l and the primary efﬂuent BOD average 160 mg/l. The 1

MG aeration tank has an MLSS concentration of 2800 mg/L and the MLVSS volatile solids

content is 75%. Calculated the F:M ratio for this plant. Solution:

F=5.5*160*8.34=7339lbs/day BOD

M=1*2800*0.75*8.34=17514lbs MLVSS

7339

17514

F:M=

= 0.41

Note: The 160 mg/l BOD concentration of the primary efﬂuent was used for the F calculation

and not 230mg/l - which is the BOD concentration of the ﬂow coming into the plant

22.24.3 Sludge Volume Index (SVI)

• SVI measures the settleability and compactibility of the secondary sludge

• It is calculated using results from the 30-minute settleability test and the MLSS concentration

•

SVI is expressed in ml/g and it is essentially the volume (ml) of 1 gram of the MLSS after 30

minutes of settling

it provides a more accurate picture of the sludge settling characteristics than settleability or

MLSS alone

50 to 120 ml/gm SVI value is considered optimal. Higher SVI values indicate sludge that

is slow to settle and not compacting well. When SVI values are approaching 200 ml/gm,

activated sludge process is considered to be "bulking".

Settled sludge volume in ml/l after 30 min

MLSS mg/l

SVI (ml/g)=

∗1000

Key Points for Solving SVI Problems

•

For the settling test MLSS is typically settled in a 1 liter settleometer. The volume of the settled

solids is therefore read as ml/L. So if for any reason a larger or smaller volume of the mixed liquor

sample is taken, the settle solids value should commensurate with the volume of the MLSS sample.

For example, if a 2 liter settleometer is used and if the solids settle to 400 ml in that settleometer,

376

Chapter 22. Water Math

the ml/L will be 400ml/2L or 200ml/L

•

For some problems, the settled solids volume is provided as a percentage (%). So if a 1-liter sett-

lometer is used and the settled solids volume is reported as 25%, it implies a settled sludge volume

of 250ml/L

Example Problem:

i. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test indicates 230 ml/L.

What is the sludge volume index?

Solution:

230ml/l

SVI=

∗1000

= 92ml/g

2500mg/l

22.24.4 Sample Math Problems

1. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the

minimum MLSS concentration, given the following:

Q = 0.4 MGD

Primary inﬂuent BOD = 250 mg/l

Primary efﬂuent BOD = 128 mg/l

Aeration tank vol. = 350,000 gallons

Clariﬁer vol = 250,000 gallons

MLSS has 80% volatile solids

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

=⇒ F : M ∝

=⇒ F:M is inversely proportional to MLSS

MLSS concentration

So to have minimum MLSS conc. F:M needs to be the maximum of the range provided

0.4∗128∗8.34

If the MLSS concentration = x: =⇒ F : M = 0.28 =

=⇒ x = 5,446mg/l MLSS

0.35∗x∗0.8

22.25 Solids Thickening Calculations

377

2. Given that an activated sludge plant with an inﬂuent ﬂow of 1.2 MGD is operated at an MCRT of 6

days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day.

Solution:

Two aeration tanks – 0.5 MG each

Two ﬁnal clariﬁers – 0.25 MG each

WAS – 7500 ppm

Final efﬂuent = 20

MLSS –3600

MLSS volatile solids content = 80%

lbs MLSS(system)

MCRT =

lbs

day

lbs

day

E f fluent_SS

WAS_SS

Step 1: Calculate the lbs MLSS (system):

lbs MLSS (system) = (2∗0.5+2∗0.25)MG∗3,600

Step 2: Calculate the lbs/day Efﬂuent_SS:

∗8.34 = 45,036 lbs

lbs

day

E f fluent_SS= 1.2MG∗20

∗8.34 = 200.2lbs

Step 3: Plug in the values in the MCRT formula:

45,036

MCRT: 6 days =

lbs

200.2

WAS_SS

day day

Step 4: Solving for WAS_SS:

lbs

day

45,036

lbs

day

WAS_SS

−200.2 = 7,306

Step 5: Solving for WAS ﬂow using the lbs formula:

lbs

day

7,306

= WAS Flow (MGD)∗7,500∗8.34

7,306

gal

day

=⇒ WAS Flow (MGD) =

= 0.116MGD = 116,000

7,500∗8.34

22.25 Solids Thickening Calculations

1. Calculate the air required (SCFM) to meet a 0.04 lb air:lb feed solids ratio for a 100 GPM WAS

ﬂow with a solids content of 6500mg/l? Assume 0.08 lbs air/SCF air. Solution:

lb air

lb solids

0.08 lbs air SCF ∗X SCF per minute

= 0.04 =

100

∗6,500

∗8.34 lbs solids

1,000,000 min

0.04∗5.421

0.08

=⇒ x SCF per minute =

= 2.7 SCFM

2. A treatment plant receives an inﬂuent ﬂow of 30 MGD with a TSS concentration of 280 mg/l. The

primary treatment removes 55% TS and the primary sludge is pumped to a 40 ft diameter gravity

378

Chapter 22. Water Math

thickener. Calculate the average solids loading to the thickener in lbs TSS/day-ft²

Solution:

(30 MGD ∗ 280∗0.55 mg/l ∗8.34)lbs TSS per day

0.785∗40²ft²

Solids loading to gravity thickener=

30.7 lbs TSS/day− ft²

22.26 Digester Calculations

22.26.1 Sludge Pumping

Example Problems:

1. A primary clariﬁer receives an average ﬂow of 12 MGD containing 280mg/L of TSS. This clariﬁer

typically removes 75% TSS and produces a 3.5% sludge and the sludge pump is rated to pump 35

cu.ft/min.

a. How many pounds of TSS is removed in the clariﬁer?

b. How many cu.ft of sludge at the given 3.5% sludge needs to be pumped per day to remove

the solids?

c. How many minutes would the sludge pump need to be operational each day to pump the

required amount of sludge - calculated from ii. above?

d. For how many minutes each hour the sludge pump should be programmed to operate (Given

the number of minutes the pump need to operate per day calculated from iii. above) ?

Solution:

a. lbs solids removed = (280∗0.75)mg/l ∗12MGD∗8.34 = 21,017 lbs solids per day

ꢁ

(

ꢁ

21,017 lbs solids x ft³sludge 7.48 ga(l sludge 0.035 lbs solids 8.34 lb sludge

(

ꢁ

(

ꢁ

(

ꢁ

∗

ꢁ

(

ft³sludge

ꢁ

(

ga(l sludge

(

ꢁ

day

1 lb sludge

ꢁ

x ft³sludge

day

21,017

0.035∗834∗7.48

ft³sludge

= 9,626

day

9,626 ft³sludge

min

275 minutes

∗

day

35 ft³sludge

day

275 minutes

day

24 hrs

11.4 minutes

∗

day

2. Calculate the lbs/day of solids removed in a primary clariﬁer treating a 5 MGD ﬂow with an aver-

age inﬂuent and efﬂuent TSS concentrations of 250 mg/l and 98 mg/l respectively. Solution:

lbs

day

lbs

= 5MGD∗(250−98)

∗8.34 = 6338

day

22.26.2 Sludge Blending

•

In digestion the feed is typically a blend of primary and secondary sludges - each of which have

different total and organic solids content.

• Thus, for blending calculations, if:

C₁and V₁is the concentration and volume (ﬂow) respectively of the primary sludge portion of the

digester feed, and

22.26 Digester Calculations

379

C₂and V₂is the concentration and volume (ﬂow) respectively of the secondary sludge feed C₃and

and V₃is the concentration and volume (ﬂow) respectively of the digester sludge feed

The sum of the mass from each of the two source sludge feeds will equal to the mass in the digester

sludge feed:

C₁* V₁+ C₂* V₂= C₃* V₃.

This equation can be manipulated algebraically to calculate anyone of the unknown values in the

equation.

Also, any of the three volume variables can be expressed as the sum or difference of the other two -

, or V₁+ V₂= V₃or V₁= V₃- V₂or V₂= V_T- V₁

Example Problem: Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1%

solids. 28,000 gal. secondary sludge at 1.3% solids. What is the combined solids concentration? If the

primary sludge is 68% VS and the secondary sludge is 63% VS, what is the VS concentration (%) in the

combined sludge?

Solution:

Combined solids concentration:

C₁∗V₁+C₂∗V₂= C₃∗V₃

C₁∗V₁+C₂∗V₂C₁∗V₁+C₂∗V₂

4.1∗15,000+1.3∗28,000

15,000+28,000

=⇒ C₃=

= 2.28%

V₁+V₂

Lbs of VS in combined sludge:

C_VS1∗V₁+C_VS2∗V₂= C_VS3∗V₃

C_VS1∗V₁+C_VS2∗V₂C_VS1∗V₁+C_VS1∗V₂

=⇒ C_VS3

V₁+V₂

4.1∗15,000∗0.68+1.3∗28,000∗0.63

15,000+28,000

=⇒ C_VS3

= 1.50%

22.26.3 Digester VS or Organic Loading

Example Problems:

1. How many pounds of TS and VS are pumped to a digester each day if the digester receives 10,000

gpd of sludge at 5% solids concentration with an average VS% of 75%?

Solution:

Digester TS loading (lbs/day)

lbs TS 10,000 gal sludge (8.34∗0.05)lbs TS)

lbs TS

day

∗

= 4,170

day

gal sludge

Digester VS loading (lbs/day)

lbs TS

day

lbs VS

lb TS

lbs VS

day

= 4,170

∗0.75

= 3,128

2. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge ﬂow. The

sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic

foot per day?

380

Chapter 22. Water Math

lbs VS

Solution:

Digester Loading

day

Digester volatile solids loading rate =

Digester volume(V) ft³

gal sludge

lbs VS

gal sludge

5000

∗(8.34∗0.06∗0.66)

lbs VS

day− ft³

day

= 0.057

(

∗37²∗27)ft³

22.26.4 Total volatile solids (VS) reduction

•

This provide a measure of the organic matter content removed and converted into digester gas in

the digester.

• Higher volatile solids reduction implies higher gas production and lower biosolids hauling costs.

• The VS reduction of the digester is provided by the Van Kleeck equation

VS_in−VS_out

Digester VS reduction(%) =

∗100

VS_in−VS_in∗VS_out

•

Digester volatile solids concentration is typically expressed as a percentage of the sludge total

solids.

• 70% VS which means that 70% of the total solids is volatile solids.

•

The value of VS_inand VS_outfor the digester VS reduction (Van Kleek) equation above should be in

fraction and not as a percentage.

A value of 0.7 should be used in the equation if the VS concentration is 70%. Likewise, as 0.525 if

the VS concentration is 52.5%.

Applying this equation to calculate the digester VS reduction if the inlet sludge VS averages 75%

and the outlet sludge is 58%?

Example Problem:

Calculate the % VS reduction in a digester given the volatile solids content of the inﬂuent sludge to

the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%

0.7−0.525

Solution: Digester VS reduction(%) =

∗100 = 53%

0.7−0.7∗0.525

22.27 Dewatering math problems

Solving Dewatering Problems

1. Calculation of solids recovery

(a) Calculate amount of solids fed to the dewatering unit

(b) Calculate the amount of solids produced as part of the dewatered cake

recovery (solids recovery rate)

2. Calculation of dewatered cake volume

22.27 Dewatering math problems

381

(a) First calculate the amount of cake solids produced in terms of weight per time.

(b) From the weight of the cake produced calculate the volume - from the cake density which is

normally given

3. Calculation of solids hauling costs or savings associated with change in cake solids content

(a) First calculate the amount of dry solids produced as part of the original wet cake solids

percent

(b) Using the value of the dry solids calculate the wet cake weight with the new cake solids

percentage

4. General formula for calculating net savings associated with change in cake solids content

(New solids %−Old Solids %)

Savings =

∗Old Cost

New solids %

So if the cake dryness goes up from 20% to 26% and currently a utility is spending $ 1,000,000 per

year for biosolids hauling and disposal, their net savings will be: (26%-%20)/26*$ 1,000,000 = $

230,769

22.27.1 Calculation of solids recovery

Example Problems:

(a) Calculate amount of solids fed to the dewatering unit

(b) Calculate the amount of solids produced as part of the dewatered cake

recovery (solids recovery rate)

22.27.2 Calculation of dewatered cake volume

(a) First calculate the amount of cake solids produced in terms of weight per time.

(b) From the weight of the cake produced calculate the volume - from the cake density which is normally

given

22.27.3 Hauling cost impact due to solids content change

(a) First calculate the amount of dry solids produced as part of the original wet cake solids percent

(b) Using the value of the dry solids calculate the wet cake weight with the new cake solids percentage

General formula for calculating net savings associated with change in cake solids content:

(New solids(%)−Old Solids(%)

Savings =

∗Old Cost

Old solids(%)

So if the average cake dryness goes up from 20% to 26% and currently this utility is spending $ 1,000,000

per year for biosolids hauling and disposal, their net savings will be:

(26%−20%)

∗1,000,000 = $300,000

20%

Example Problems

1. 12,000 ft³of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The

centrifuge yields 37 yd³of 26% of dewatered cake with a density of 73 lb/ft³. Calculate the solids

capture rate.

Solution:

gal (8.34∗0.028lbs TS)

= 20,960lbs TS

lbs TS feed to centrifuge = 12,000ft³sludge∗7.48

∗

ft³

gal sludge

ft³(73lbs∗0.26 TS)

lbs TS feed from centrifuge = 37yd³sludge ∗ 27

∗

= 18,961lbs TS

yd³

ft³sludge

382

Chapter 22. Water Math

18,961lbs solids produced by centri fuge

20,960lbs solids fed from digester

solids capture rate =

∗100 = 90.4%solids capture

2. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at

68 lbs/ ft³. How long would it take to ﬁll a 3 yd³bin

Solution:

cake TS produced −lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS

∗

∗0.9

lb sludge

min

galllon

12.61lbs TS

min

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

0.927 ft³

27 ft³

yd³

Time required to fill the bin =

∗3yd³∗

= 75min

Chapter Assessment

Practice Problems - Fractions

1. Convert 22 into a fraction

2. Express 10ft, 6in into fraction

3. Express 10ft, 6in into decimal

Practice Problems - Decimals and Powers of Ten

1. Write the equivalent of 10,000,000 as a power of ten

2. Find the product of 3.4564∗10²

3. Find the product of 534.567∗10⁻²

165.93

4. Find the value of

10⁻²

5. Find the value of 0.023∗10⁴

Solutions:

1. 10⁷

2. 345.64

3. 5.34567

4. 16,593

5. 230

384

Chapter 22. Water Math

Practice Problems - Rounding and Signiﬁcant Digits

Round the following to the nearest hundredths (the second place after the decimal).

A. 2.4568

B. 27.2534

C. 128.2111

D. 364.8762

E. 354.777777

F. 34.666666

G. 67.33333

Solution:

A. 2.46

B. 27.25

C. 128.21

D. 364.88

E. 354.78

F. 34.67

G. 67.33

Round the following to the nearest tenths (the ﬁrst place after the decimal).

A. 2.4568

B. 27.2534

C. 128.2111

D. 364.8762

E. 354.777777

F. 34.666666

G. 67.33333 Solution:

A. 2.5

B. 27.3

C. 128.2

D. 364.9

E. 354.8

F. 34.7

G. 67.3

Round the following answers off to the most signiﬁcant digit.

Problem

Accurate Answer

25.1+26.43 = 51.53

128.456−121.4 = 7.056

85−7.92432 = 77.07568

8.564+5 = 13.564

22.27 Dewatering math problems

385

Problem

Accurate Answer

26.34×124.34567 = 3,275.26495

23.58×34.251 = 807.63858

12,453/13.9 = 895.8992805755

12,457.92×3 = 37,373.76

Practice Problems - Averages

1. Find the average of the following set of numbers:

0.2

0.1

0.3

0.2

0.4

0.6

0.1

0.3

2. The chemical used for each day during a week is given below. Based on these data, what was the

average lb/day chemical used during the week?

Monday

Tuesday

Wednesday

Thursday

Friday

92 lb/day

93 lb/day

98 lb/day

93 lb/day

89 lb/day

93 lb/day

97 lb/day

Saturday

Sunday

3. The average chemical use at a plant is 77 lb/day. If the chemical inventory is 2800 lbs, how many

days supply is this?

4. A well pumped for 45 days. The beginning meter reading was 7,456,400 and 45 days later the same

meter was 15,154,400. What was the average ﬂow in gallons per day?

Practice Problems - Percentage

1. 25% of the chlorine in a 30 -gallon vat has been used. How many gallons are remaining in the vat?

2. The annual public works budget is $147,450. If 75% of the budget should be spent by the end of

September, how many dollars are to be spent? How many dollars will be remaining?

3. A 75 pound container of calcium hypochlorite has a purity of 67%. What is the total weight of the

calcium hypochlorite?

386

Chapter 22. Water Math

4. 3/4 is the same as what percentage?

5. A 2% chlorine solution is what concentration in mg/L ?

6. A water plant produces 84,000 gallons per day. 7,560 gallons are used to backwash the ﬁlter. What

percentage of water is used to backwash?

7. The average day winter demand of a community is 14,500 gallons. If the summer demand is esti-

mated to be 72% greater than the winter, what is the estimated summer demand? Demand - When

related to use, the amount of water used in a period of time. The term is in reference to the "de-

mand" put onto the system to meet the need of customers.

8. The master meter for a system shows a monthly total of 700,000 gallons. Of the total water, 600,000

gallons were used for billing. Another 30,000 gallons were used for ﬂushing. On top of that,

15,000 gallons were used in a ﬁre episode and an estimated 20,000 gallons were lost to a main

break that was repaired that same day. What is the total unaccounted for water loss percentage for

the month?

9. Your water system takes 75 coliform tests per month. This month there were 6 positive samples.

What is the percentage of samples which tested positive?

Total volume to be pumped

Time =

Pump flow rate

ꢀ

7.48gal

(0.785∗110²∗25) ft³∗

ꢀ

ft³

=⇒

= 1,251 min

ꢀ

1420gal

min

ꢀ

Practice Problems - Ratio and Proportion

1. It takes 6 gallons of chlorine solution to obtain a proper residual when the ﬂow is 45,000 gpd. How

many gallons will it take when the ﬂow is 62,000 gpd?

2. A motor is rated at 41 amps average draw per leg at 30Hp. What is the actual Hp when the draw is

36 amps? C.

3. If it takes 2 operators 4.5 days to clean an aeration basin, how long will it take three operators to do

the same job?

4. It takes 3 hours to clean 400 feet of collection system using a sewer ball. How long will it take to

clean 250 feet?

5. It takes 14 cups of HTH to make a 12% solution, and each cup holds 300 grams. How many cups

will it take to make a 5% solution?

6. A bike travelling at 5 miles/hr completes a journey in 40 minutes. How long would the same jour-

ney take if the speed was increased to 8 miles/hr?

Solution

1. The gallons chlorine and ﬂow are directly related.

Thus,

6∗62,000

45,000

=⇒ X =

= 8.3gallons

45,000 62,000

22.27 Dewatering math problems

387

2. The amp draw and Hp are directly related.

This

41 36

30∗36

=⇒ X =

= 26.3Hp

3. The number of operators and the days to clean are inversely related.

Thus,

2∗4.5

2∗4.5 = 3∗X =⇒ X =

= 3days

4. The hours to clean and the length of system cleaned are directly proportional.

Thus,

3∗250

=⇒ X =

= 1.9hours

400 250

400

5. The cups of HTH and percentage HTH solution are directly proportional.

Thus,

14∗5

=⇒ X =

= 5.8cups

6. The bike speed and time to complete the journey are inversely related.

Thus,

5∗40

5∗40 = 8∗X =⇒ X =

= 25min

Practice Problems - Area and Volume

1. A 60-foot diameter tank contains 422,000 gallons of water. Calculate the height of water in the

storage tank.

2. What is the volume of water in ft³, of a sedimentation basin that is 22 feet long, and 15 feet wide,

and ﬁlled to 10 feet?

3. What is the volume in ft³of an elevated clear well that is 17.5 feet in diameter, and ﬁlled to 14 feet?

4. What is the area of the top of a storage tank that is 75 feet in diameter?

5. What is the area of a wall 175ft. in length and 20ft. wide?

6. You are tasked with ﬁlling an area with rock near some of your equipment. 1 Bag of rock covers

250 square feet. The area that needs rock cover is 400 feet in length and 30 feet wide. How many

bags do you need to purchase?

7. A circular clearwell is 150 feet in diameter and 40 feet tall. The Clearwell has an overﬂow at 35

feet. What is the maximum amount of water the clearwell can hold in Million gallons rounded to

the nearest hundredth?

8. A sedimentation basin is 400 feet length, 50 feet in width, and 15 feet deep. What is the volume

expressed in cubic feet?

9. A clearwell holds 314,000ft³of water. It is 100ft in diameter. What is the height of the clearwell?

10. A treatment plant operator must ﬁll a clearwell with 10,000ft³of water in 90 minutes. What is the

rate of ﬂow expressed in GPM?

11. A water tank has a capacity of 6MG. It is currently half full. It will take 6 hours to ﬁll. What is the

ﬂow rate of the pump?

12. A clearwell with the capacity of 2.5MG is being ﬁlled after a maintenance period. The ﬂow rate is

388

Chapter 22. Water Math

2,500 GPM. The operator begins ﬁlling at 7 AM. At what time will the clearwell be full?

13. A chemical feed pump with a 6-inch bore and a 6-inch stroke pumps 60 cycles per minute. Find the

pumping rate in gpm.

14. Determine the ﬂow capacity of a pump in gpm if the pump lowers the water level in a 6 -foot

square wet well by 8 inches in 5 minutes.

15. How much paint will it take for a single coat of the top and sidewalls of the storage tank that is

100-feet in diameter and 30-feet tall, if one gallon of paint covers 200 square feet?

a. 86 gallons

b. 96 gallons

c. 106 gallons

d. 116 gallons

e. 126 gallons

16. Under like conditions, how much more water would an 8-inch pipe carry than a 4-inch pipe?

a. 2 times

b. 3 times

c. 4 times

d. not enough information given

Solution:

Volume

1. Volume = Surface area * height =⇒ height =

Surface area

Diameter=60’

Volume=422,000 gal

Height=h’

ꢂ

422,000 gꢀal∗

ꢀ

7.48 gꢀal

ꢀ

=⇒ height =

= 101 ft

0.785∗60²ft²

Practice Problems - Flow and Velocity

1. A rectangular channel 3 ft. wide contains water 2 ft. deep ﬂowing at a velocity of 1.5 fps. What is

the ﬂow rate in cfs?

2. Flow in an 8-inch pipe is 500 gpm. What is the average velocity in ft/sec? (Assume pipe is ﬂowing

full)

3. A pipeline is 18” in diameter and ﬂowing at a velocity of 125 ft. per minute. What is the ﬂow in

gallons per minute?

4. The velocity in a pipeline is 2 ft./sec. and the ﬂow is 3,000 gpm. What is the diameter of the pipe in

inches?

5. Find the ﬂow in a 4-inch pipe when the velocity is 1.5 feet per second.

6. A 42-inch diameter pipe transfers 35 cubic feet of water per second. Find the velocity in ft/sec.

7. A plastic ﬂoat is dropped into a channel and is found to travel 10 feet in 4.2 seconds. The channel

22.27 Dewatering math problems

389

is 2.4 feet wide and 1.8 feet deep. Calculate the ﬂow rate of water in cfs.

8. The ﬂow velocity of a 6-inch diameter pipe is twice that of a 12-inch diameter pipe if both are

carrying 50 gpm of water. True or false?

9. What should the ﬂow meter read in gpm if a 4-inch diameter main is to be ﬂushed at a velocity of

4.6 fps?

10. The velocity through a channel is 4.18 fps. If the channel is 4 feet wide by 2 feet deep by 10 feet

long, what is the ﬂow rate in gpm?

11. What is the average ﬂow velocity in ft/sec for a 12-inch diameter main carrying a daily ﬂow of

2.5mgd ?

Solution:

1. Solution:

sec

ft³

sec

Q = V ∗A =⇒ Q = 1.5

∗(3∗2)ft²= 9

2. Solution:

Q = V ∗A

min

ꢂ

ft ∗

ꢁ

min

500gallon

ꢁ

60sec

∗

ꢀ

ꢁ

ꢀ

7.48gal

ꢀ

=⇒ V =

=⇒ V

= 3.2 ft/s

ꢀ

ꢁ

ꢂ

0.785∗

3. Solution:

The diameter of the pipe is 4 inches. Therefore, the radius is 2 inches. Convert the 2 inches to feet.

= 0.6667ft

A = π ×r²

A = π ×(0.167ft)²

A = π ×0.028ft²

A = 0.09ft²

Q = V×A

Q = 1.5ft/sec×0.09ft²

Q = 0.14ft/3sec(cfs)

390

Chapter 22. Water Math

Practice Problems - Unit Conversions

Convert the following:

1. Convert 1000 ft³to cu. yards

2. Convert 10 gallons/min to ft³/hr

3. Convert 100,000 ft³to acre-ft.

4. Find the ﬂow in gpm when the total ﬂow for the day is 65,000 gpd.

5. Find the ﬂow in gpm when the ﬂow is 1.3cfs.

6. Find the ﬂow in gpm when the ﬂow is 0.25cfs.

7. The ﬂow rate through a ﬁlter is 4.25 MGD. What is this ﬂow rate expressed as gpm?

8. After calibrating a chemical feed pump, you’ve determined that the maximum feed rate is 178 mL/

minute. If this pump ran continuously, how many gallons will it pump in a full day?

9. A plant produces 2,000 cubic foot of water per hour. How many gallons of water is produced in an

8-hour shift?

Solution

1. Solution:

cu.yards

27 ft³

1000 ft³∗

= 37cu.yards

2. Solution:

ꢁ

10 gallons

ft³

60min 80.2 ft³

ꢁ

∗

ꢁ

min

7.48gallons

ꢁ

3. Solution:

acre− ft

100,000ft³∗

= 2.3acre− ft

ꢁ

43,560 ft − ft

ꢁ

Note: From the conversion table: acre = 43,560 ft²

Thus, acre-ft = 43,560 ft²-ft or 43,560 ft³

4. Solution:

65,000gpd

1,440 min/day

= 45gpm

5. Solution:

cfs 448gpm

1.3

= 582gpm

1cfs

6. Solution:

cfs 448gpm

0.25

= 112gpm

1cfs

7. Solution:

Flowrate,gpm =

Flow rate, gpd

1440 min/day

Note: We are assuming that the ﬁlter operated uniformly over that 24 hour period.

gal

4.25

∗1,000,000

day

Flowrate,gpm =

8. Solution:

= 2,951 gpm

min

1440

day

22.27 Dewatering math problems

391

2000 ft³7.48 gallons 60 hr

119,680 gallons

∗

ft³

hꢀr

shift

ꢀ

Practice Problems - Concentration

1. What is the concentration in mg/l of 4.5% solution of that substance.

2. How many lbs of salt is needed to make 5 gallons of a 2,500mg/l solution

Solution

1. 45,000 mg/l

2. Applying pounds formula: lbs salt =

1,000,000

∗2,500∗8.34 = 0.14lbs

Practice Problems - Density and Speciﬁc Gravity

1. What is the speciﬁc gravity of a 1 ft³concrete block which weighs 145 lbs?

2. What is the speciﬁc gravity of a chlorine solution if 1 (one) gallon weighs 10.2lbs?

3. How much does each gallon of zinc orthophosphate weigh (pounds) if it has a speciﬁc gravity of

1.46?

4. How much does a 55 gallon drum of 25% caustic soda weigh (pounds) if the speciﬁc gravity is

1.28?

Practice Problems - Detention Time

1. A ﬂocculation basin is 7 ft deep, 15 ft wide, and 30 ft long. If the ﬂow through the basin is 1.35

MGD, what is the detention time in minutes?

2. A tank has a diameter of 60 feet with an overﬂow depth at 44 feet. The current water level is 16

feet. Water is ﬂowing into the tank at a rate of 250 gallons per minute. At this rate, how many days

will it take to ﬁll the tank to the overﬂow?

3. How long will it take to ﬁll a 50 gallon hypochlorite tank if the ﬂow is 5gpm ?

4. Find the detention time in a 45,000 gallon reservoir if the ﬂow rate is 85gpm.

5. If the fuel consumption to the boiler is 35 gallons per day. How many days will the 500 gallon tank

last.

6. The sedimentation basin in a water plant contains 5,775 gallons. What is the detention time if the

ﬂow is 175gpm.

Solution

1. Solution:

392

Chapter 22. Water Math

7’ deep

1.5 MGD

15’ wide

30’ long

gal

ft³

(30∗15∗7)ft³∗7.48

DT =

= 25min

gal

day

1,350,000

∗

day 1440min

2. Solution:

250 gpm

Diameter=60’

Overﬂow level = 44’

Current level = 16’

7.48gallons

0.785∗60²∗(44−16) ft³∗

ft³

Volume

Flow

Fill time =

= 1.6 days

gallons 1440 min

250

∗

min

day

3. Solution:

50gal

5gal/min

4. Solution:

45,000gal

DT =

= 10 min

529 min

60 min/hr

DT =

= 529 min or

= 14.3 days

= 33 min

= 8.8hrs

85gal/min

5. Solution:

500 gal

DT =

35gal/ day

6. Solution:

5,775gal

DT =

175gal/min

22.27 Dewatering math problems

393

Practice Problems - Pounds Formula

1. A water treatment plant operates at the rate of 75 gallons per minute. They dose soda ash at 14

mg/L. How many pounds of soda ash will they use in a day?

2. A water treatment plant is producing 1.5 million gallons per day of potable water, and uses 38

pounds of soda ash for pH adjustment. What is the dose of soda ash at that plant?

3. A water treatment plant produces 150,000 gallons of water every day. It uses an average of 2

pounds of permanganate for iron and manganese removal. What is the dose of the permanganate?

4. A water treatment plant uses 8 pounds of chlorine daily and the dose is 17 mg/l. How many gallons

are they producing?

5. An operator mixes 40 lb of lime in a 100-gal tank containing 80 gal of water. What is the percent of

lime in the slurry?

6. A treatment plant has a maximum output of 30MGD and doses ferric chloride at 75 mg/L. How

many pounds of Ferric Chloride does the plant use in a day?

7. A treatment plant uses 750 pounds of alum a day as it treats 15MGD. What was the dose rate?

8. A treatment plant operates at 1,500 gallons a minute and uses 500 pounds of alum a day. What is

the alum dose?

Solution:

1. Solution:

lbs/day

14 mg/l

75 GPM

8.34

lbs

day

= Flow

∗Concentration

∗8.34

ꢁ

gallons

ꢁ

lbs

day

min

lbs

= 75

∗1440

∗

∗250

∗8.34 = 225

l day

ꢁ

min

day 1,000,000 gallons

ꢁ

2. Solution:

38 lbs/day

mg/l

1.5 MGD

8.34

394

Chapter 22. Water Math

lbs

day

= Flow

∗Concentration

∗8.34 =⇒ Concentration

Concentration

Flow

∗8.34

lbs

day

= 3

1.5

∗8.34

day

3. Solution:

38 lbs/day

mg/l

1.5 MGD

8.34

lbs

day

= Flow

∗Concentration

∗8.34 =⇒ Concentration

Concentration

Flow

∗8.34

lbs

day

= 3

ꢁ

Gallons

ꢁ

150,000

∗

∗8.34

(

day

1,000,000 Gallons

4. Solution:

8 lbs/day

17 mg/l

MGD

8.34

lbs

day

= Flow

∗Concentration

∗8.34

lbs

day

lbs

day

=⇒ Flow

= 0.056425

Concentration

∗8.34 17

∗8.34

MG 1,000,000 Gallons

0.056425

∗

= 56,425 Gallons

day

5. Solution:

lbs = Volume(MG)∗Concentration

∗8.34

22.27 Dewatering math problems

395

38 lbs/day

mg/l

1.5 MGD

8.34

lbs

40 lbs

1,000,000 gallons

=⇒ Concentration

Volume(MG)∗8.34

80 gallons∗

∗8.34

ꢁ

Practice Problems - Temperature Conversion

1. Convert 22^◦C into degree Fahrenheit.

2. Convert 56^◦C into degree Celsius.

Preliminary Treatment

1. On an average a 12.5 yd. load of grit is hauled to the landﬁll once every 20 days. Plant ﬂow aver-

ages 12.5 MGD. Calculate the rate of grit collection in ft3/MG.

Correct Answer(s):

a. 1.4

2. A grit channel is 2 feet 4 inches wide. When wastewater is ﬂowing 8 inches deep in this channel,

the velocity is found to be 1.6 ft per second. Calculate ﬂow in MGD.

a. 1.55 MGD

*b. 1.60 MGD

c. 1.90 MGD

d. 2.50 MGD

3. A grit channel is 2 feet 4 inches wide. When wastewater is ﬂowing 8 inches deep in this channel,

the velocity is found to be 1.6 ft per second. Calculate ﬂow in MGD.

a. 1.55 MGD

*b. 1.60 MGD

c. 1.90 MGD

d. 2.50 MGD

4. At a wastewater treatment plant which receives a ﬂow rate of 650,000 gallons per day, a total of 50

cubic feet of grit was removed for the month. Calculate the rate of grit removal assuming 30 days

in a month.

396

Chapter 22. Water Math

Solution:

GritRemoval _MG= 50

ft³

ꢁ

d¨ay

gal

ꢁ^month

ꢀ

∗

_650,000gal∗1,000,000_MG= 2.6

ꢁ

ꢁ^month

30days

ꢀ

5. On an average a 12.5 yd. load of grit is hauled to the landﬁll once every 20 days. Plant ﬂow aver-

ages 12.5 MGD. Calculate the rate of grit collection in ft³/MG.

Solution:

ꢀ

ꢁ

ft³

27 ft

d¨ay

^12.5ꢀ^yꢀd

GritRemoval

∗

= 1.4

20days

12.5MG

^yꢀ^d

ꢀ

6. On an average, 2 inches of grit is collected and removed every day in a 2.2 feet wide, 205 feet long

grit channel. Knowing the average ﬂow through that grit channel is 10 MGD calculate the rate of

grit collection in ft³/MG

Solution:

75.16 ft

day

ꢀ

day

ꢀ

ft³

75.16ft

day

Grit volume accumulated: ₁₂ft ∗(2.2∗205) ft²=

Grit Collection:

= 7.5

10MG

7. What is the grit production rate (cu. ft/MG) if 46 cu. ft of grit is removed every ﬁve days and the

daily plant ﬂow averages 2 MGD.

a. 12 cu. ft/MG

b. 23 cu. ft/MG

*c. 4.6 cu. ft/MG

d. 9.2 cu. ft/MG

Solution:

ꢀ

ꢁ

ft³

46ft

ꢁ

d¨ay

Grit Production Rate

∗

= 4.6

ꢁ

days 2MG

8. A grit chamber is 2 feet 4 inches wide. When wastewater is ﬂowing 8 inches deep in this channel,

the ﬂow velocity is found to be 1.6 ft per second. Calculate ﬂow in MGD.

Solution:

Q = V ∗A

Q = 1.6

ꢀ

∗

ꢁ

12 12

ft³

∗

ft²= 2.49

ft³(1440∗60)s

gal MG

ꢀ

Q = 2.49

∗7.48

= 1.61MGD

day

ft³

gal

ꢀ

9. A steel bin is used to collect the daily accumulation of grit for disposal in a land ﬁll. The bin is 8

ft long, 5.25 ft. wide, and 6 feet deep. When the bin is three-quarters full it is hauled away. This

occurs on the average every 10 days. Average ﬂow to the plant is

3.75 MGD. What is the rate of grit collection in units of cubic feet per MG of ﬂow?

Correct Answer(s):

22.27 Dewatering math problems

397

a. 4.9

10. A ﬂow controlled grit chamber has two channels which are each 2.5 feet wide, and 20 feet long.

Only one of these channels is currently in service and it receiving a ﬂow of 4.5 MGD. The wastewa-

ter is ﬂowing 1.5 feet deep in this channel. Calculate the velocity of this

ﬂow. Should we put the other channel in service?

Correct Answer(s):

11. Calculate the ﬂow, in gpd, that would pass through a grit chamber 2 feet wide, at a depth of 6

inches, with a velocity of 1 ft /sec.

Correct Answer(s):

a. 646272.0

Primary Treatment

1. A clariﬁer has a TSS removal efﬁciency of 50%. If the inﬂuent TSS concentration is 220 mg/L,

how many lbs/day of TSS are removed if the ﬂow is 10 MGD. Also, how many cu. ft of sludge is

pumped if the sludge has a TS concentration of 5%.

lbs solids removed = (220∗0.50)mg/l ∗10MGD∗8.34 = 9,174lbs solids per day

(

ꢁ

(

ꢁ

(

ft³sludge 9,174 lbs solids

1 lb sludge

ga(l sludge

ft³sludge

day

(

ꢁ

(

ꢁ

∗ ∗ ∗

ꢁ

(

0.05 lbs solids 8.34lb sludge

(

ꢁ

= 2,941

(

ꢁ

(

ꢀ

day

7.48 gal

ꢀ

2. A community has a total ﬂow of 15 MGD which is passed through a primary treatment plant which

removes 60% of the TSS and 35% of the BOD. The average strength of the inﬂuent is 400 mg/l

TSS and 275 mg/l BOD. If the total solids of the raw sludge is 5%, how many cu. ft of sludge is

pumped daily? lbs solids removed = (400∗0.60)mg/l∗15MGD∗8.34 = 30,024lbs solids per day

(

ꢁ

(

ꢁ

(

ft³sludge 30,924 lbs solids

1 lb sludge

ga(l sludge

ft³sludge

day

(

ꢁ

(

ꢁ

∗ ∗ ∗

ꢁ

(

0.05 lbs solids 8.34lb sludge

(

ꢁ

= 9,626

(

ꢁ

ꢀ

day

7.48 gal

ꢀ

3. How many lbs of solids are removed daily by a primary clariﬁer treating a 6 MGD ﬂow if the

average inﬂuent TSS concentration is 300 mg/l and the clariﬁer TSS removal efﬁciency is 67%?

As the removal efﬁciency is 67%, 0.67 * 300 mg/l = 201 mg/l solids are removed.

The total lbs removed can be calculated using the lbs formula.

lbs solids

mg SS

lbs solids

day

= 6MGD∗201

∗8.34 = 10,058

day

4. Calculate the primary clariﬁer inﬂuent solids concentration if its outlet concentration is 60 mg/l and

the known clariﬁer removal efﬁciency is 75%?

Actual inlet (X)

Actual outlet

Actual inlet (X)

100

100−Removal e f ficiency

100

= 4

100−75

=⇒ Actual inlet (X) = 4∗60 = 240mg/l

5. A circular clariﬁer receives a ﬂow of 11 MGD. If the clariﬁer is 90 ft. in diameter and is 12 ft. deep,

what is: a) the hydraulic/surface loading rate, b) weir overﬂow rate, and c) clariﬁer detention time

in hours?

Solution:

398

Chapter 22. Water Math

a) Hydraulic/surface loading rate:

10 gal

11MG

ꢀ

ꢁ

ꢀ

∗

day

gpd

= 1,730gpd/ ft²

ꢀ

ꢀ ₂

Clarifier hydraulic loading

0.785∗90 ft

b) Weir overﬂow rate:

10 gal

11MG

ꢀ

ꢁ

ꢀ

∗

day

gpd

ꢀ

_3.14∗9ꢀ_0ft

Weir over flow rate

= 38,924gpd/ft

c) Clariﬁer detention time:

Clarifier volume(cu. ft or gal)

In fluent flow (cu. ft or gal)/hr)

Clarifier detention time (hr) =

(0.785∗90 ∗15)ft

Clarifier detention time (hr) =

= 2hrs

day

gal

₂ꢀ_4hrs

∗

7.48gal

¹⁰ꢀ

11MG

ꢀ

∗

day

ꢀ

6. A clariﬁer has a TSS removal efﬁciency of 50%. If the inﬂuent TSS concentration is 220 mg/L,

how many lbs/day of TSS are removed if the ﬂow is 10 MGD. Also, how many cu. ft of sludge is

pumped if the sludge has a TS concentration of 5%.

Solution:

lbs solids removed = (220∗0.50)mg/l ∗10MGD∗8.34 = 9,174lbs solids per day

(

ꢁ

(

ꢁ

(

ft³sludge 9,174 lbs solids

1 lb sludge

ga(l sludge

ft³sludge

day

(

ꢁ

(

ꢁ

∗ ∗ ∗

ꢁ

(

0.05 lbs solids 8.34lb sludge

(

ꢁ

= 2,941

(

ꢁ

(

ꢀ

day

7.48 gal

ꢀ

7. If a clariﬁer has a capacity of 0.25 MG, what is the detention time in hours if it receives a ﬂow of 3

MGD

Solution:

Clarifier volume(MG)

Clarifier detention time (hr) =

In fluent flow (MG/hr)

0.25MG

Clarifier detention time (hr) =

= 2hrs

day

3MG

ꢀ ꢀ

ꢀ

∗

24hrs

day

ꢀ

8. If a clariﬁer has a capacity of 0.25 MG, what is the detention time in hours if it receives a ﬂow of 3

MGD

Solution:

Clarifier volume(MG)

Clarifier detention time (hr) =

In fluent flow (MG/hr)

0.25MG

Clarifier detention time (hr) =

= 2hrs

day

3MG

ꢀ ꢀ

ꢀ

∗

24hrs

day

ꢀ

9. At a 2.5 MGD wastewater treatment plant the primary clariﬁer has a detention time of 2 hours.

How many gallons does this clariﬁer hold?

Solution:

Clarifier volume(gal)

Clarifier detention time (hr) =

In fluent flow (gal/hr)

=⇒ Clarifier volume(gal) = Clari fier detention time (hr)∗Influent flow (gal/hr)

ꢀ

ꢁ ꢀ

ꢁ

gal

day

=⇒ Clarifier volume(gal) = 2 hrs ∗ 2.5∗10⁶

∗

= 208,333 gals

day 24 hrs

10. What is the surface loading rate (gal/(day-sq.ft) of a 15 MGD ﬂow in a 105 ft diameter primary

sedimentation tank operating at water depth of 20 ft.

Solution:

22.27 Dewatering math problems

399

15MG 10⁶gal

∗

ꢀ

ꢁ

gpd

ft²

day

Clarifier hydraulic loading

= 1,733gpd/ft²

0.785∗105²ft²

11. If a 90 ft diameter primary clariﬁer operating at water depth of 20 ft is treating a 12MGD ﬂow,

calculate the surface loading rate (gal/(day-sq.ft).

Solution:

12MG 10⁶gal

∗

ꢀ

ꢁ

gpd

ft²

day

0.785∗90²ft²

Clarifier hydraulic loading

= 1,887gpd/ft²

12. What is the weir overﬂow rate (gpd/ft) when treating a 15 MGD ﬂow in a 105 ft diameter primary

sedimentation tank operating at water depth of 20 ft.

Solution:

(0.785∗105²∗20) ft³

Clarifier detention time (hr) =

= 2.1hrs

ft³

day

ꢀ

15MG 10 gal

ꢀ

∗

ꢀ

24hrs

7.48gal

day

ꢀ

13. A circular clariﬁer receives a ﬂow of 3.2 MGD. If the diameter is 70 ft, what is the weir overﬂow

rate?

Solution:

3.2MG 10⁶gal

∗

ꢀ

ꢁ

gpd

day

3.14∗70 ft

Weir over flow rate

= 14,559gpd/ ft

14. How many lbs/day of solids are removed in a clariﬁer treating a 6 MGD ﬂow if the average inlet

concentration is 320 mg/l and its average outlet concentration is 80 mg/l

Solution:

Applying pounds formula:

lbs solids

= 6MGD∗(320−80)^mg∗8.34 = 120,096

lbs solids removed

day

15. At a 2.5 MGD wastewater treatment plant the primary clariﬁer has a detention time of 2 hours.

How many gallons does this clariﬁer hold?

16. What is the detention time (hrs) in a 105’ diameter primary sedimentation tank operating at water

depth of 20’ when treating a 15 MGD ﬂow (Ans: 2.1)

17. What is the surface loading rate (gal/(day-sq.ft) of a 15 MGD ﬂow in a 105 ft diameter primary

sedimentation tank operating at water depth of 20 ft. (Ans: 1,733)

18. What is the weir overﬂow rate (gpd/ft) when treating a 15 MGD ﬂow in a 105 ft diameter primary

sedimentation tank operating at water depth of 20 ft. (Ans: 45,496)

19. A circular clariﬁer receives a ﬂow of 3.2 MGD. If the diameter is 70 ft, what is the weir overﬂow

rate? (Ans: 14,559)

20. A clariﬁer has an inﬂuent suspended solids concentration of 420 mg/L. If the suspended solids

efﬁciency is 62%, how many mg/L suspended solids are removed? (Ans: 260)

21. Given the following for a primary sedimentation tank:

TSS removal efﬁciency = 63%

Efﬂuent TSS concentration = 95 mg/L

400

Chapter 22. Water Math

Calculate the inﬂuent TSS concentration (mg/L)

(Ans: 257)

22. What is the clariﬁer inﬂuent TSS if its outlet concentration is 60 mg/l and the known clariﬁer

removal efﬁciency is 75%?

*a. 240 mg/l

b. 300 mg/l

c. 80 mg/l

d. 120 mg/l

23. What is the clariﬁer removal efﬁciency if the inlet and outlet concentrations are 300 mg/l and 60

mg/l respectively?

a. 20%

b. 30%

*c. 80%

d. 72%

24. A rectangular sedimentation tank is 85 feet long, 35 feet wide, and 14 feet deep including 3 feet of

freeboard. Flow to this tank is 2.3 MGD. Calculate the surface loading to this tank in gpd per ft2.

a. 318 gpd/ft2

*b. 773 gpd/ft2

c. 845 gpd/ft2

d. 1932 gpd/ft2

25. What is the detention time in a 40 ft diameter primary clariﬁer operating with a 12 ft water depth,

when treating a 1.5 MGD ﬂow?

a. 0.24 hours

b. 0.60 hours

*c. 1.8 hours

d. 2.3 hours

e. 11.0 hours

26. How many lbs/day of solids are removed in a clariﬁer treating a 6 MGD ﬂow if the average inlet

concentration is 320 mg/l and its average outlet concentration is 80 mg/l

a. 16,000 lbs/day

*b. 1,210 lbs/day

c. 4,000 lbs/day

d. 9,000 lbs/day

27. At a 5 MGD wastewater ﬂow, a primary clariﬁer has a detention time of 1 hour. How many gallons

does this clariﬁer hold?

a. 104,000 gallons

*b. 208,000 gallons

c. 250,000 gallons

d. 500,000 gallons

e. 5,000,000 gallons

28. Percent efﬁciency of total solids or BOD removal is calculated using the following formula: (In-

Out*100)/(In-(In*Out))

a. True

*b. False

29. The only function of ﬂights in a rectangular sedimentation tank is to push the settled sludge towards

22.27 Dewatering math problems

401

the hopper

a. True

*b. False

30. A settling basin has a length of 60 ft., width of 20 ft., and depth of 12 ft. At a ﬂow rate of 60 mgd,

what is the retention time of this basin?

a. 15 minutes.

b. 30 minutes.

*c. 1.1 hours.

d. 2.3 hours.

31. Calculate the detention time for a sedimentation tank that is 48 feet wide, 210 feet long and 9 feet

deep with a ﬂow of 5 MGD.

*a. 3.25 hours.

b. 3.63 hours.

c. 5.65 hours.

d. 5.82 hours.

32. Calculate the weir loading for a sedimentation tank that has an outlet weir 480 ft. long and a ﬂow

of 5 MGD.

a. 9220 gpd/ft.

b. 9600 gpd/ft.

c. 9920 gpd/ft.

*d. 10,420 gpd/ft.

33. If a primary clariﬁer consistently operates at 30 % efﬁciency and produces an efﬂuent which aver-

ages 140 mg/l BOD, what is the inﬂuent BOD?

a. 100 mg/l

b. 125 mg/l

c. 175 mg/l

*d. 200 mg/l

e. 225 mg/l

34. At a 2.5 MGD wastewater treatment plant the primary clariﬁer has a detention time of 2 hours.

How many gallons does this clariﬁer hold?

a. 104,000 gallons

*b. 208,000 gallons

c. 250,000 gallons

d. 500,000 gallons

e. 5,000,000 gallons

35. A 3.1 MGD ﬂow with a 190 mg/l TSS concentration is treated in a primary clariﬁer which averages

55% removal efﬁciency. Calculate the pounds/day TSS in the primary efﬂuent.

a. 2445 lbs/day

*b. 2211 lbs/day

c. 3800 lbs/day

d. 4200 lbs/day

36. A 75 ft long, 25 ft wide and 10 ft deep sedimentation basin receives a ﬂow of 1.6 MGD. What is

the detention time in hours.

a. 7 hrs

b. 1.7 hrs

402

Chapter 22. Water Math

*c. 2.1 hrs

d. 4 hrs

37. A sludge pump is set to pump 5 minutes each hour. It pumps at the rate of 35 gpm. How many

gallons of sludge are pumped each day?

a. 120 gpd

b. 175 gpd

c. 840 gpd

*d. 4200 gpd

38. A primary clariﬁer has an inﬂuent suspended solids concentration of 250 mg/L. If the suspended

solids removal efﬁciency is 60%, what is the primary efﬂuent suspended solids concentration

a. 10 mg/L

*b. 100 mg/L

c. 50 mg/L

d. 150 mg/L

39. A treatment plant receives a ﬂow of 3.5 MGD. If the clariﬁer is 100 ft. long, 30 ft. wide, and 15 ft.

deep, what is the surface loading rate?

a. 78 gal/ft/day

b. 700 gal/ft/day

*c. 1,170 gal/ft/day

d. 1,500 gal/ft/day

e. 4,500 gal/ft /day

40. A settling basin has a length of 60 ft, width of 20 ft, and depth of 12 ft At a ﬂow rate of 60 MGD,

what is the detention time of this basin?

a. 15 minutes

b. 30 minutes

*c. 1.1 hours

d. 2.3 hours

41. If a clariﬁer has a capacity of 0.25 MG, what is the detention time in hours if it receives a ﬂow of 3

MGD

a. 1.8 hrs

*b. 2.0 hrs

c. 2.25 hrs

d. 2.50 hrs

e. 3.00

22.27 Dewatering math problems

403

Trickling Filter

1. The inﬂuent to a trickling ﬁlter plant is 200 mg/L and the efﬂuent BOD is 20 mg/L. What is the

BOD removal efﬁciency (

2. The ﬂow to a trickling ﬁlter is 1.33 MGD. If the primary efﬂuent has a BOD concentration of231

mg/Land the trickling ﬁlter efﬂuent has a BOD concentration of 83 mg/l, how many pounds of

BOD are removed? (Ans: 1642 lbs/day)

3. A 80-ft diameter trickling ﬁlter with a media depth of 7 ft receives a ﬂow of 2,180,000 gpd. If the

BOD concentration of the primary efﬂuent is 139 mg/L, what is the organic loading on the trickling

ﬁlter in lbs BOD/day/1000 cu ft? (Ans: 72 lbs/day-1000 ft3)

4. A standard-rate ﬁlter, 90 ft in diameter, treats a primary efﬂuent ﬂow of 540,000 gpd. If the recir-

culated ﬂow to the trickling ﬁlter is 120,000 gpd, what is the hydraulic loading rate on the ﬁlter in

gpd/sq ft? (Ans: 104)

5. A 90 feet diameter trickling ﬁlter treats a 540,000 gpd primary efﬂuent ﬂow. If the recirculated ﬂow

to the trickling ﬁlter is 120,000 gpd, what is the hydraulic loading on the trickling ﬁlter in gpd/ft2.

*a. 104 gpd/ft2

b. 2 gpd/ft2

c. 3 gpd/ft2

d. 4 gpd/ft2

6. A trickling ﬁlter plant operating at a recirculation ratio of 1 receives a raw wastewater ﬂow of 2

MGD. This means that the ﬂow being applied to the ﬁlter would be:

*a. 4mgd.

b. 1mgd

c. 3mgd

d. 2mgd

e. 6mgd

7. What is the hydraulic loading of a trickling ﬁlter with a 100-foot diameter, if it receives a ﬂow of

4.0 MGD?

a. 480 GPD/ft2

*b. 510 GPD/ft2

c. 540 GPD/ft2

d. 570 GPD/ft2

e. 600 GPD/ft2

8. A 12 MGD primary clariﬁer efﬂuent ﬂow is fed to a trickling ﬁlter. If the total ﬂow to the trickling

ﬁlter including the recirculated ﬂow is 17 MGD, what is the recirculation ratio?

(Ans: 0.42)

9. A trickling ﬁlter 100 ft. in diameter and 6 ft. deep receives a ﬂow of 0.5 mgd with a BOD of 150

mg/L. What is the BOD loading in pounds per day per 1000 cubic feet of ﬁlter media?

a. 11.9.

*b. 13.3.

c. 26.8.

d. 133.

10. Calculate the organic load in lbs of BOD/day/1000 cu. ft. for a trickling ﬁlter 50 ft in diameter and

eight (8) ft deep. Given a ﬂow of 250,000 gpd, and a 150 mg/l BOD concentration of the primary

efﬂuent feed to the trickling ﬁlter.

404

Chapter 22. Water Math

a. 2 lbs/day/1000 cu. ft.

b. 5 lbs/day/1000 cu. ft.

*c. 20 lbs/day/1000 cu. ft.

d. 50 Ibs/day/1000 cu. ft.

11. Calculate the pounds of BOD per day entering the trickling ﬁlter.

DATA: Raw wastewater ﬂow is 1.5 MGD.

Raw wastewater BOD is 150 mg/L.

There is a 30*a. 1600 lbs/day

b. 880 lbs/day

c. 870 lbs/day

d. 560 lbs/day

12. A trickling ﬁlter wastewater treatment plant receives a ﬂow 1.95 MGD. Calculate the organic

loading to this plant if it has a 135 ft diameter trickling ﬁlter with a 5 foot media depth and has a

primary efﬂuent BOD concentration of 110 mg/L.

a. 0.5 lbs BOD/1,000 ft³/day.

b. 2.7 lbs BOD/1,000 ft³/day.

*c. 25 lbs BOD/1,000 ft³/day.

d. 39 lbs BOD/1,000 ft³/day.

e. 44 lbs BOD/1,000 ft³/day.

13. At a 2.1 MGD wastewater treatment plant the inﬂuent suspended solids concentration to the pri-

mary clariﬁer is 240 mg/l. The primary sludge contains 3.2gallons of primary sludge should be

pumped per day?

a. 63,000 gal/day ·

b. 32,200 gal/day

c. 8,200 gal/day

*d. 7,550 gal/day

14. Calculate the pounds of BOD per day entering the trickling ﬁlter

DATA:

Raw wastewater ﬂow is 15 MGD

Raw wastewater BOD is 150 mg/L

There is a 30

a. 560 lbs/day

b. 870 1 bs/ day

c. 880 1 bs /day

*d. 1600 lbs/day

15. Calculate the organic load in lbs of BOD/day/1000 cu ft for a trickling ﬁlter 50 ft in diameter and

8 feet deep, a BOD design loading of 50 lbs/day/1000 cu ft, a ﬂow of 250,000 gpd, and a primary

efﬂuent BOD to ﬁlter of 150 mg/L

a. 2 lbs/day/1000 cu ft

b. 5 lbs/day/1000 cu ft

*c. 20 lbs/day/1000 cu ft

d. 50 lbs/day/1000 cu ft

16. What is the trickling ﬁlter hydraulic loading if the inﬂuent ﬂow is 3 MGD and the recirculation rate

is 4:1?

a. 3 mgd

22.27 Dewatering math problems

405

b. 6mgd

c. 9mgd

d. 12mgd

*e. 15 mgd

f. None of the above

17. Calculate the organic load in lbs of BOD/day/1000 cu. ft. for a trickling ﬁlter 50 ft in diameter and

eight (8) ft deep. Given a ﬂow of 250,000 gpd, and a 150 mg/l BOD concentration of the primary

efﬂuent feed to the trickling ﬁlter.

a. 2 lbs/day/1000 cu. ft.

b. 5 lbs/day/1000 cu. ft.

*c. 20 lbs/day/1000 cu. ft.

d. 50 Ibs/day/1000 cu. ft.

18. The desired trickling ﬁlter recirculation ratio is 1.4. If the primary efﬂuent ﬂow is 4.4 MGD what is

the trickling ﬁlter efﬂuent ﬂow that needs to be recirculated.

(Ans: 6.2)

Stabilization Ponds

1. A 40 acre pond receives a 0.6 MGD ﬂow. If the pond is operated at a depth of 4ft, what is the

detention time (days) of this pond. (Ans: 87)

2. The ﬂow to a pond is 750 gpm. The pond width is 200 ft and length is 450 ft. If the BOD in the

pond inﬂuent is 240mg/L, what is the organic loading to this pond in lbs-BOD/day/acre? (Ans:

1046)

3. Calculate the depth the liquid level must be lowered in the fall below the normal 3-foot depth to

prepare for winter and provide 180 day storage. DATA: Wastewater stabilization ponds total area of

13.8 acres (600,000 sq. ft.) Average winter wastewater ﬂow to the ponds is 100,000 gpd. Maximum

liquid storage depth is 5 feet. (Assume no loss of liquid by percolation or evaporation and that the

ponds will ﬁll to the maximum storage level of 5 feet.)

a. 0.1 ft.

b. 1.0 ft.

*c. 2.0 ft.

d. 4.0 ft.

4. Compute the lagoon’s detention time (days) DATA: Surface area= 6.0 acres Average depth= 3.0

feet Average evaporation exceeds precipitation by 0.4 in/day Average daily ﬂow (inﬂuent)= 0.25

MGD 1 acre (area)= 43,560 sq ft 1 cu ft contains 7.5 gal

a. 24

*b. 32

c. 37

d. None of the above

5. A pond is 600 ft. long and 150 ft. wide. If the pond is 6 ft. deep, what is the volume in acre ft?

(Ans: 13.9)

6. A wastewater pond has a detention time of 30 days and is operated at a pond depth of 4 ft. What is

the hydraulic loading on the pond?

a. 7.5 in/day

*b. 1.6 in/day

406

Chapter 22. Water Math

c. 0.63 in/day

d. 0.41 in.day

e. 013 in/day

7. The volume of an aerated pond is 700,000 cubic feet. Flow to the pond averages .050 MGD. How

many days detention time does the pond provide?

a. 10.5 days.

b. 14 days.

*c. 105 days.

d. 140 days.

8. Assume a community of 1,100 people has a lagoon. The ﬂow averages 80,000 gallons per day. The

ﬂow is

a. 14 gallons per capita.

*b. 72 gallons per capita.

c. 80 gallons per capita.

d. 140 gallons per capita.

9. What is the organic loading on a lagoon (lbs./day/acre) if the inﬂuent BOD averages 164 mg/l and

the ﬂow averages 0.67 MGD. The lagoon has two (2) identical square cells with each side of 800 ft.

a. 29 lbs./ac.

b. 23 lbs./ac.

*c. 31 lbs./ac.

d. 52 lbs./ac.

10. A stabilization pond is 1100 ft. long, 600 ft wide, and is operated at a depth of 5 ft. It receives a

ﬂow of 500,000 gpd and which has an inﬂuent BOD of 185mg/L. Using this information do the

following:

(a) Convert the ﬂow to the pond in units of acre-ft/day (Ans. 1.53 acre-ft/day)

(b) Find the area of this pond in units of acres. (Ans. 540,000 ft3 and 13.9 acre-ft)(Ans. 15.2

acres)

(d) Calculate the pond detention time in days. (Ans. 44.9)

(e) Calculate the hydraulic loading to the pond in units of inches per day. (Ans. 1.2 in/day)

(f) Calculate the organic loading to the pond (lbs of BOD/day/acre) (Ans. 50.8 lbs BOD/day-

acre)

Solution:

500,000gal

ft³

acre− ft

(a)

∗

= 1.53

day

7.48gal 43,560 ft³

day

acres

(b) (1,100∗600) ft²∗

= 15.2acres

43,560ft²

acres− ft

= 75.8acre− ft

43,560ft³

volume

flow

7.48gal

(d) DT =

= (1,100∗600∗5) ft³∗

= 49.4days

500,000fracgalday ft³

day

Pond depth (in)

5∗12in

= =

(e) Pond hydraulic loading rate

Volume

Flow

49.4

Pond detention time

1.2

day

22.27 Dewatering math problems

lbs BOD per day

407

(f) Organic loading =

area (acres)

0.5MGD ∗ 185mg/l ∗ 8.34 43,560ft²

50.9lbs BOD

day−acre

=⇒

∗

1,100∗600ft²

acre

11. What is the surface area of a pond that is 4 feet deep, if it holds 30 million gallons?

Solution:

Pond Volume = Sur face Area∗Depth =⇒ 30MG = Sur face Area∗4ft

acre− ft

30MG∗3.069

=⇒ Sur face Area (acres) =

= 23 acre

4ft

12. The inﬂuent ﬂow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L

in the raw wastewater. How many lbs of suspended solids are sent to the pond daily?

Solution:

lbsSS

day

gal 24hrs

lbs SS

∗8.34 = 284

day

= 10000

∗

∗ ∗142

day 1000000gal

13. A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in acres?

Solution:

(260∗80) ft²∗

acre

= 0.48acre

43,560 ft²

14. A pond has a volume of 1,800,000 ft³. If the ﬂow to the pond is 425 gpm, what is the pond deten-

tion time in days?

Solution:

volume

flow

day

1440min

7.48gal

DT =

= 1,800,000ft³∗

∗

= 22days

ft³

gal

425

min

15. Find pond hydraulic loading in inches/day when the depth of the pond is 6 ft. and the detention

time is 30 days.

Solution:

flow

Hydraulic Loading (HL) =

area

vol

Detention time (DT) =

=⇒ flow =

flow

Substituting for ﬂow in the HL formula above:

vol

area

vol

area∗DT

pond depth

vol

area

HL =

=⇒ HL =

= pond depth

Pond depth (in)

6∗12 inches

2.4in

day

Pond hydraulic loading rate =

Volume

Flow

30 days

Pond detention time

16. The ﬂow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond inﬂuent is

170mg/L, what is the organic loading to this pond in lbs BOD/day/acre?

Solution:

lbs BOD per day 7.2MGD ∗ 170mg/l ∗ 8.34 43,560ft²

4,624lbs BOD

day−acre

Organic loading =

∗

area (acres)

0.785∗350²ft²

acre

17. The ﬂow to a pond is 750,000 gpd. If the pond diameter is 100 ft and the BOD in the pond inﬂuent

is 300 mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans. 10,412 lbs

BOD/day-acre)

Solution:

lbs BOD per day

0.75MGD ∗ 300mg/l ∗ 8.34 43,560ft²

Organic loading =

∗

area (acres)

0.785∗100²ft²

acre

10,413lbs BOD

day−acre

408

Chapter 22. Water Math

18. A 40 acre wastewater treatment pond receives a ﬂow of 0.6 MGD. If the pond is operated at a depth

of 4ft. What is the detention time of this pond?

Solution:

Volume

Flow

(40∗4)acre− ft

gal

ft³

acre− ft

day 7.48gal 43,560ft³

Pond detention time =

= 87 days

0.6∗10⁶

∗

19. A 2.5 acre stabilization pond is operated at a depth of ﬁve (5) ft. What is the pond detention time if

the ﬂow to the pond is 18,000 cu. ft/day?

Solution:

Volume

Flow

(2.5∗5)acre− ft

ft³acre− ft

Pond detention time =

= 30 days

18,000

∗

day 43,560ft³

20. The inﬂuent to a pond is 1,200 gpm with a BOD concentration of 180mglL in the raw wastewater.

How many lbs/day of BOD are sent to the pond daily? (Ans. 2,600 lbs BOD/day)

21. The inﬂuent ﬂow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L

in the raw wastewater. How many lbs of suspended solids are sent to the pond daily? (Ans. 284 lbs

SS/day)

22. A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in ft2 and acres? (Ans. 20,800

ft2)

23. A pond has a volume of 1,800,000 ft3. If the ﬂow to the pond is 425 gpm, what is the pond deten-

tion time in days? (Ans. 22 days)

24. Find pond detention time in days given the following:

• Flow to pond = 250 gpm

• Pond length = 125 ft .

• Pond width = 40ft.

• Pond depth = 7ft.

(Ans. 0.7 days)

25. Find hydraulic loading in inches/day for a pond given the following:

• Pond depth = 12ft.

• Pond volume = 1,400,000ft3

• Pond ﬂow = 1,000,000gal/day

Solution:

gal

ft³

day 7.48gal

1,000,000

∗

day

Flow

Area

day

Pond hydraulic loading rate

=⇒

∗12 = 13.8

1,400,000ft³

12 ft

Note: The area of the pond was found by dividing the volume (1,000,000 ft³) by the pond depth (12ft)

26. The ﬂow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond inﬂuent is

170mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans. 0.3 lbs BOD/day-

acre)

27. The ﬂow to a pond is 200,000 gph. The pond width is 650 ft and the length is 300 ft. If the BOD in

the pond inﬂuent is 265mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans.

2,370 lbs BOD/day-acre)

28. The inﬂuent ﬂow to a pond is 600 gpm with a BOD concentration of 150mg/L in the raw wastewa-

ter. How many lbs of BOD are sent to the pond daily? (Ans. 1081 lbs BOD/day)

29. The inﬂuent ﬂow to a pond is 50,000 gallons/hour with a suspended solids concentration of 125mg/L

in the raw wastewater. How many lbs of suspended solids are sent to the pond daily? (Ans. 1,251

22.27 Dewatering math problems

409

lbs TSS/day)

30. The ﬂow to a pond is 60 gpm with a COD concentration of 240mg/L in the pond inﬂuent. How

many lbs/day of COD are sent to the pond daily? (Ans. 173 lbs COD/day)

31. A pond is 100 ft. long and 210 ft wide. What is the area of this pond in ft2 and acres?(Ans. 21,000

ft2 and 0.48 acres)

32. A pond has a diameter of 80ft. What is the area of this pond in ft2 and acres? (Ans. 5,024 ft2 and

0.12 acres)

33. A pond is 450 ft. long and 450 ft. wide. If the pond is 3 ft. deep, what is the volume in ft3 and acre

ft? (Ans. 607,500 ft3 and 13.9 acre-ft)

34. A pond has a diameter of 230 ft. and is 6 ft. deep. What is the volume in ft3 and acre ft of this

pond? (Ans. 249,159 ft3 and 5.7 acre-ft)

35. A pond has a diameter of 50 ft. and is 8 ft deep. What is the volume of this pond in ft3 and acre ft?

(Ans. 15,700 ft3 and 0.36 acre-ft)

36. A pond has a volume of 2,800,000 ft3. If the ﬂow to the pond is 500 gpm, what is the detention

time in days? (Ans. 29 days)

37. A 75 acre facultative pond is operated at a depth of 5.5 feet. Flow to this pond averages 2.2 MGD.

The inﬂuent pond BOD averages,195 mg/L. Calculate the following:

(a) The detention time (Ans. 61 days)

(b) The hydraulic loading (Ans. 1.08 in/day)

38. How deep must a 60 acre facultative pond be operated in order to have a detention time of 45 days.

Flow to the pond is 2.0 MGD. (Ans. 4.6 ft)

39. Find pond detention time in days.

• Flow to pond = 300,000 gpm

• Pond width = 450 ft.

• Pond length = 500 ft.

• Pond depth = 5 ft.

(Ans. 28 days)

40. A 75 acre pond receives an inﬂuent ﬂow of 5 MGD. If the operating level of the pond is to be raised

from 4.0 ft to 5.5 ft, how long will it take for the pond level to rise to the new discharge level (Ans.

7.3 days)

41. Find the pond detention time in days if the ﬂow to the pond is 600 gpd, and the pond is 450ft wide,

500 ft long and operating at a depth of 5 ft.

42. The ﬂow to a pond is 500 gpm with a COD concentration of 240mg/L in the pond inﬂuent. How

many lbs/day of COD are present to the pond daily?

43. The ﬂow to a pond is 864,000 gal/day. The suspended solids concentration in the pond inﬂuent is

380mg/L. How many lbs/ day of suspended solids are sent to the pond daily?

44. A 5 acre stabilization pond is operated at a depth of ﬁve (5) ft. What is the pond detention time if

the ﬂow to the pond is 30,000 cu. ft/day?

Solution:

Volume

Flow

(5∗5)acre− ft

ft³acre− ft

Pond detention time =

= 36.3 days

30,000

∗

day 43,560ft³

45. What is the hydraulic loading rate (in/day) on a 20 acre pond if it receives a ﬂow of 3.3 acre-ft/day.

a. 1 in/day

410

Chapter 22. Water Math

b. 12 in/day

c. 0.165 in/day

*d. 2 in/day

46. It is found that at a wastewater treatment pond system 35 days are required to bring the pond sys-

tem to its operations depth of 3.5 feet. Assuming that the pond system was empty when the ﬁlling

of the pond was initiated and that a constant incoming ﬂow rate ﬁlls the pond, what would be the

hydraulic loading on this pond system ?

a. 0.47 inches/day

*b. 1.2 inches/day

c. 2.1 inches/day

d. 3.5 inches/day

e. 10.0 inches/day

Activated Sludge

1. The aeration tank of a conventional activated sludge plant contains 9300 lbs of MLSS. Lab tests

show that the MLSS are 79% volatile solids. The primary efﬂuent contains 2900 lbs of BOD.

Calculate the F to M ratio for this wastewater treatment plant. (Ans: 0.4)

2. How many lbs of solids are in a 400,000 gallon aeration tank if the suspended solids concentration

is 1200 mg/l? (Ans: 4003.0)

3. Operational data is given below for a conventional activated sludge treatment plant:

Inﬂuent ﬂow: 2.5 mgd

Inﬂuent BOD: 220 mg/L

Inﬂuent TSS: 240 mg/L

Primary BOD removal efﬁciency: 30%

RAS SS: 8,500 mg/L

RAS VSS: 6,630 mg/L

RAS ﬂow: 160%

Aeration tank volume: 1.8 MG

Secondary clariﬁer volume: 0.8 MG

MLSS: 3,600 mg/L

MLVSS: 2,800 mg/L

WAS ﬂow: 35,000 gpd

AS Efﬂuent TSS: 25 mg/L

AS Efﬂuent BOD: 19 mg/L

Settleability results: 60 min= 300 ml/L

Settleability results: 30 min= 320 ml/L

1) Calculate the MCRT (8 points)

2) Calculate the F/M Ratio (8 points)

3) Calculate the SVI (8 points)

4) Calculate the aeration basin detention time

4. Calculate the F to M ratio given the following.

Plant ﬂow- 0.8 MGD

Aerator vol- 200,000 gal.

Clariﬁer Volume-175,000 gal

22.27 Dewatering math problems

411

Primary eff BOD- 120 mg/L

MLVSS conc. -1950 mg/L

(Ans: 0.25)

5. Quantify the amount (lbs) of MLVSS in the two (2) aeration tanks if each tank is 40 ft. long, 15

ft. wide and operating at a water depth of 12 ft. The MLSS is 2500 mg/l and is found to be 69%

volatile. (Ans: 1728 lbs)

6. The aeration tank of a conventional activated sludge plant contains 12,000 lbs of MLSS. Lab tests

show that the MLSS are 79% volatile solids. The primary efﬂuent contains 3800 lbs of BOD.

Calculate the F to M ratio for this wastewater treatment plant. (Ans: 0.4)

7. An activated sludge plant with the following parameters treats 5 MGD ﬂow.

Aeration tank volume: 0.25 MG

Clariﬁer volume: 0.15 MG

MLSS: 3000 mg/l

RAS concentration: 6500 mg/l

Secondary efﬂuent TSS: 16 mg/l

MCRT: 6 days

1. Calculate the lbs MLSS in the aeration system (Aeration tank + Clariﬁer)

2. Calculate the lbs TSS/day in the ﬁnal efﬂuent - 5 points

3. At the given MCRT, calculate the WAS solids removed from the system (lbs/day)

4. Using the given value of WAS concentration, calculate the gallons per day of WAS

8. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test for a 1-litre MLSS

samples indicates 230 ml of settled solids. What is the sludge volume index? (Ans: 92)

9. An activated sludge process treating a 3 MGD ﬂow has a 1 MG volume aeration basin and a 0.25

MG volume ﬁnal clariﬁer. The MLSS is 2000 mg/l and contains 82% volatile solids. If the sec-

ondary efﬂuent suspended solids concentration is 20mg/l and the WAS concentration is 7500 mg/l

and is wasted at a rate of 0.06 MGD. Calculate the MCRT. (Ans: 4.9)

10. The desired F/M ratio at a particular activated sludge plant is 0.5 lbs BOD/lb MLVSS. If the 3

MGD primary efﬂuent ﬂow has a BOD of 165 mg/l, how many lbs of MLVSS should be main-

tained in the aeration tank. (Ans: 8,257)

11. Operational data is given below for a conventional activated sludge treatment plant:

Inﬂuent ﬂow: 0.500 mgd

RAS SS: 12600 mg/L

RAS VSS: 9450 mg/L

Inﬂuent BOD: 199 mg/L

Inﬂuent TSS: 221 mg/L RAS ﬂow: 160%

Aeration tank volume: 350,000 gal WAS ﬂow: 6433 gpd

Secondary clariﬁer volume: 150,000 gal

MLSS: 4200 mg/L Efﬂuent TSS: 45 mg/L

MLVSS: 3150 mg/L Efﬂuent VSS: 36 mg/L

Settleability results: 30 min= 360 ml/L Efﬂuent BOD: 20 mg/L

Settleability results: 60 min= 330 ml/L

Calculate the MCRT

Answer: 21 days

Calculate the F/M Ratio

Answer: 0.09

412

Chapter 22. Water Math

Calculate the aeration tank detention time

Answer: 6.5 hrs

Calculate the SVI

Answer: 88 mL/g

12. Calculate the F to M ratio given the following.

Plant ﬂow- 0.8 MGD

Aerator vol- 200,000 gal.

Clariﬁer Volume-175,000 gal

Primary eff BOD- 120 mg/L

MLVSS conc. -1950 mg/L

(Ans: 0.25)

13. An operator ran a 30 minute settling test on a mixed liquor sample using a 1000 ml graduated

cylinder. The mixed liquor settled to 22% of the cylinder volume. The lab determined that the

MLSS concentration was 3380 mg/l. What was the Sludge Volume Index for the mixed liquor?

Correct Answer: 65.0

14. The desired F/M ratio is .35lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,

how many lbs of MLVSS should be maintained in the aeration tank?

a. 3430 lbs

b. 420 lbs

*c. 6000 lbs

d. 735 lbs

15. A sludge settleability test shows a reading 220 ml after 30 minutes of settling in a one liter gradu-

ated cylinder. Lab testing on this mixed liquor shows a MLSS concentration of 1850 mg/L and a

MLVSS concentration of 1440 mg/L. Calculate SVI for this mixed liquor sample.

*a. 119

b. 108

c. 153

d. 138

16. Given the following data, calculate the mean cell residence time (MCRT) .

DATA: Inﬂuent ﬂow = 3.5 MGD. MLSS = 2850 mg/L. Waste sludge ﬂow = 0.08 MGD. Total

secondary system volume = 2.0 MG. Waste activated sludge suspended solids conc -6000mg/L.

Final efﬂuent suspended solids = 25 mg/L.

a. 6 days.

b. 8 days.

*c. 10 days.

d. 12 days.

17. What is the Sludge Volume Index given the following: MLSS = 2800 mg/L; MLVSS 240 mg/l;

Settled Volume after 30 minutes = 250 ml/L

a. 0.09 ml/g

*b. 89 ml/g

c. 0.11 ml/g

d. 113 ml/g

18. Given the following data, calculate the suspended solids required to maintain a food to microorgan-

ism (F/M) ratio of 0.3

DATA:

22.27 Dewatering math problems

413

Daily ﬂow = 10 MGD.

Average primary efﬂuent BOD = 150mg/L

Aeration tank capacity = 200,000 gal

a. 2100 mg/L

*b. 2500 mg/L

c. 3300 mg/L

d. 4200 mg/L

19. Given the following data, calculate the mean cell residence time (MCRT)

DATA:

Inﬂuent ﬂow= 35 MGD

MLSS = 2850 mg/L

Waste sludge ﬂow = 0.08 MGD

Total secondary system volume= 20 MG

Waste activated sludge suspended solids conc. = 6000 mg/L

Final efﬂuent suspended solids = 25 mg/L

a. 6 days

b. 8 days

*c. 10 days

d. 12 days

20. Calculate the waste sludge pumping rate if you want to waste 2,000 pounds per day with a concen-

tration of return sludge at 5000 mg/L

a. 29 gpm

*b. 33 gpm

c. 35 gpm

d. 37 gpm

21. Calculate the return activated sludge (RAS) concentration

DATA:

SVI = 140

SSV(30) = 350

Inﬂuent ﬂow = 25 MGD

MLSS concentration = 2500 mg/L

a. 8,300 mg /L

*b. 7,100 mg /L

c. 4,150 mg/L

d. 2,800 mg/L

22. Calculate the return activated sludge (RAS) concentration.

DATA:

SVI = 140 SSV = 350 30

Inﬂuent ﬂow= 2.5 MGD

MLSS concentration= 2500 mg/L.

a. 8300 mg /l.

*b. 7100 mg/l.

c. 4150 mg/l.

d. 2800 mg/l.

23. Given the following data, calculate the mean cell residence time (MCRT):

414

Chapter 22. Water Math

DATA: Inﬂuent ﬂow= 3.5 MGD.

MLSS = 2850 mg/L.

Waste sludge ﬂow= 0.08 MGD.

Total secondary system volume= 2.0 MG.

Waste activated sludge suspended solids conc.: 6000 mg/L.

Final efﬂuent suspended solids= 25 mg/L.

a. 6 days.

*b. 8 days.

c. 10 days.

d. 12 days.

24. Calculate the return activated sludge (RAS) concentration.

DATA:

SVI = 140 SSV = 350 30

Inﬂuent ﬂow= 2.5 MGD

MLSS concentration= 2500 mg/L.

a. 8300 mg /l.

*b. 7100 mg/l.

c. 4150 mg/l.

d. 2800 mg/l.

25. Calculate the return activated sludge (RAS) concentration.

DATA:

SVI = 140 SSV = 350 30

Inﬂuent ﬂow= 2.5 MGD

MLSS concentration= 2500 mg/L.

a. 8300 mg /l.

*b. 7100 mg/l.

c. 4150 mg/l.

d. 2800 mg/l.

26. Given the following data, calculate the mean cell residence time (MCRT) .

DATA: Inﬂuent ﬂow = 3.5 MGD. MLSS = 2850 mg/L. Waste sludge ﬂow = 0.08 MGD. Total

secondary system volume = 2.0 MG. Waste activated sludge suspended solids conc -6000mg/L.

Final efﬂuent suspended solids = 25 mg/L.

a. 6 days.

b. 8 days.

*c. 10 days.

d. 12 days.

27. At an activated sludge wastewater treatment plant receiving 3.25 MGD, the ﬁnal efﬂuent suspended

solids concentration averages 21.2 mg/L. What would the calculated MCRT value be when the

aeration basin carries 2,050 mg/L MLSS and wastes 0.0550 MGD. The waste activated sludge

has a concentration of 7,980 mg/L. The aeration tank has a volume of 1.00 MG and the secondary

clariﬁer has an operational volume of 0.250 MG.

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

22.27 Dewatering math problems

415

MLSS in aeration tank (lbs) = 1∗2050∗8.34 = 17,097lbs

MLSS in clarifier (lbs) = 0.25∗2050∗8.34 = 4,274.3lbs

SS e f fluent (lbs/day) = 3.25MGD∗21.2mg/l ∗8.34 = 574.6lbs/day

SS WAS (lbs/day) = 0.055MGD∗7,980mg/l ∗8.34 = 3,660.4lbs/day

17,097.6+4,274.3

Plugging in the values calculated above: MCRT(days) =

574.6+3,660.4

= 4.8 = 5 days

28. Calculate the MCRT of an activated sludge plant given the following information.

Plant ﬂow- 4.25 MGD

WAS conc-7980 mg/l

Waste ﬂow- 0.055 MGD

RAS conc.- 7980 mg/l

Aeration tank vol-1MG

Clariﬁer vol- 0.25 MG

Final eff TSS conc. - 21.2 mg/l

MLSS conc.- 2050 mg/l

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 1∗2050∗8.34 = 17097lbs

MLSS in clarifier (lbs) = 0.25∗2050∗8.34 = 4274.3lbs

SS e f fluent (lbs/day) = 4.25MGD∗21.2mg/l ∗8.34 = 751.4lbs/day

SS WAS (lbs/day) = 0.055MGD∗7980mg/l ∗8.34 = 3660.4lbs/day

17097.6+4274.3

Plugging in the values calculated above: MCRT(days) =

= 4.8 = 5days

751.4+3660.4

416

29. Calculate the MCRT given the following.

Chapter 22. Water Math

Plant ﬂow - 1.8 MGD

MLSS conc - 2800 mg/l

WAS ﬂow - 0.04 MGD

MLVSS conc. - 2190 mg/l

Aerator vol - 0.3 MG

Reactor vol. - 0.2 MG

RAS conc. - 8150 mg/l

Efﬂuent SS conc.-18 mg/l

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 0.3∗2800∗8.34 = 7005.6lbs

MLSS in clarifier (lbs) = 0.2∗2800∗8.34 = 4670.4lbs

SS e f fluent (lbs/day) = 1.8MGD∗18mg/l ∗8.34 = 270.2lbs/day

SS WAS (lbs/day) = 0.04MGD∗8150mg/l ∗8.34 = 2718.8lbs/day

7005.6+4670.4

270.2+2718.8

Plugging in the values calculated above: MCRT(days) =

= 3.9 = 4days

22.27 Dewatering math problems

417

30. In an aeration tank, the MLSS is 2650 mg/l and recorded 30-minute settling test indicates 221 ml/L.

What is the sludge volume index?

Solution:

Settled sludge volume in ml/l after 30 min

MLSS mg/l

SVI (ml/g)=

∗1000

221ml/l

2650mg/l

SVI=

∗1000

= 83ml/g

31. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,

how many lbs of MLVSS should be maintained in the aeration tank?

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

2100

=⇒ 0.35 =

=⇒ x = 6000lbs MLVSS

32. A sludge settleability test shows a reading 220 ml after 30 minutes of settling in a one liter gradu-

ated cylinder. Lab testing on this mixed liquor shows a MLSS concentration of 1850 mg/L and a

MLVSS concentration of 1440 mg/L. Calculate SVI for this mixed liquor sample.

*a. 119

b. 108

c. 153

d. 138

Solution:

Settled sludge volume in ml/l after 30 min

MLSS mg/l

SVI (ml/g)=

∗1000

220ml/l

1,850mg/l

SVI=

∗1000

= 119ml/g

33. Calculate F/M ratio based on the following data:

Secondary inﬂuent BOD - 156 mg/l

Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep

Inﬂuent ﬂow - 0.65 MGD

MLSS - 3600 mg/l

MLSS average % volatile - 72%

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

(lbs) MLVSS in the aeration tank

F : M =

156∗0.65∗8.34

= 0.06

7.48gal

3600∗0.72∗4∗(30∗70∗10)ft³∗

∗

∗8.34

1000000gal

ft³

418

Chapter 22. Water Math

34. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the

minimum MLSS concentration, given the following:

Q = 0.4 MGD

Primary inﬂuent BOD = 250 mg/l

Primary efﬂuent BOD = 128 mg/l

Aeration tank vol. = 350,000 gallons

Clariﬁer vol = 250,000 gallons

MLSS has 80% volatile solids

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

=⇒ F : M ∝

=⇒ F:M is inversely proportional to MLSS

MLSS concentration

So to have minimum MLSS conc. F:M needs to be the maximum of the range provided

0.4∗128∗8.34

If the MLSS concentration = x: =⇒ F : M = 0.28 =

=⇒ x = 5446mg/l MLSS

0.35∗x∗0.8

35. Given that an activated sludge plant with an inﬂuent ﬂow of 1.2 MGD is operated at an MCRT of 6

days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day.

Two aeration tanks – 0.5 MG each

Two ﬁnal clariﬁers – 0.25 MG each

WAS = 7500 ppm

Final efﬂuent = 20

MLSS volatile solids content = 80%

MLSS =3600

36. Your plant currently is running at an MCRT of 5 days. You want to adjust things so that the plant

is operating at an MCRT of 9 days. What is the new volume in gallons you should waste everyday

to achieve this? Given the following information: Aeration tank volume of 750,000 gal. MLSS of

2400 mg/l. Plant efﬂuent SS of 7 mg/l, and a WAS concentration of 6500 mg/l. Inﬂuent ﬂow 1.0

MGD. [Ans. 29,292 gallons]

37. Your engineer is upgrading your pond by providing surface aerators. She suggests you need two

7-1/2 horsepower surface aerators with an oxygen transfer rate of 1.6 pounds of oxygen per horse-

power per hour. Assume only 80% of the oxygen transfer will reduce the BOD 1:1. What is the

minimum number of hours these time-clocked aerators will run to remove 85% of the inﬂuent

BOD, which averages 295 pounds a day? [Ans. 13.1 hrs]

38. Given the following information calculate MCRT for your WWTP. The plant is an extended aera-

tion system with an aeration tank that has a volume of 66800 cu. ft. The MLSS is 1950 mg/l. The

ﬂow to the WWTP is 1.5 MGD and your efﬂuent TSS is 9 mg/l. The WAS is 7000 mg/l and you

waste 11990 gallons each day. [Ans. 10 days]

22.27 Dewatering math problems

419

39. Given a system that has 25,000 pounds of MLSS in it and the desired quantity is 23,500 pounds of

MLSS, how many gallons of sludge must be wasted if the RAS and WAS concentration is 7,200

mg/l?[Ans. 24,980 gallons]

40. Your aeration tank is 80 feet long, 40 feet wide and has 10.5 feet of mixed liquor depth. The

MLVSS in the aeration tank is 1,850 mg/l. You are experiencing biological bulking and have read

2 pounds of chlorine per 1,000 pounds of MLVSS should control this form of bulking. How many

pounds of chlorine are added to the RAS portion of the RAS/WAS splitter box to achieve this

process control strategy?[Ans. 7.8 lbs Cl2]

41. An extended aeration plant is as follows: 0.28 MGD of 210 mg/l inﬂuent BOD, aeration basin

volume 0.32 MG. The MLVSS is 4,450 mg/l (81% volatile) and the RAS is 9,200 mg/l. (a) What is

the F/M? [Ans. 0.051 F:M] (b) How many pounds of suspended solids must be wasted to obtain an

F/M of .06? [1785.6 lbs MLSS] (c) What is the new MLVSS in mg/l? [Ans. 3781 mg/l MLSS]

42. Plant data: WAS Rate – 9 gpm Flow – 1.5 MGD MLSS – 2,350 mg/l Inﬂuent BOD – 210 mg/l

RAS/WAS – 5,875 mg/l Inﬂuent TSS – 186 mg/l 30 min. Settling test – 320 ml/L Primary efﬂuent

BOD – 143 mg/l MLSS is 85% volatile solution Primary efﬂuent TSS – 86 mg/l Efﬂuent BOD – 18

mg/l Aeration basins (2) Total Volume 468,750 gal Efﬂuent TSS – 13 mg/l a). Calculate the F/M

ratio: [Ans. 0.23] b). Calculate the sludge age: [Ans. 8.54days] c). Calculate MCRT: [Ans. 11.5

days] d). Calculate the SVI: [Ans. 136]

43. The 1.5 MGD inﬂuent to a treatment plant has an average BOD of 280 mg/L. 32% of this BOD

is removed in the primary sedimentation tank. The primary efﬂuent ﬂows into an aeration tank

containing 7000 lbs of MLSS with 79 % volatile matter. Calculate this plant’s F to M ratio.

Inﬂuent BOD concentration to the AS basin: 280

∗(1−0.32) = 190.4

1.5MGD∗190.4

∗8.34

= 0.43

7000∗0.79

44. Given that an activated sludge plant with an inﬂuent ﬂow of 1.2 MGD is operated at an MCRT of 6

days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day.

Two aeration tanks – 0.5 MG each

Two ﬁnal clariﬁers – 0.25 MG each

WAS – 7500 ppm

Final efﬂuent = 20

MLSS volatile solids content = 80%

MLSS –3600

lbsMLSS(system)

MCRT=

lbs

day

lbs

day

E f fluentSS+

WASSS

lbs MLSS (system)= (2∗0.5+2∗0.25)MG∗3600

∗8.34 = 45036lbs

lbs

day

E f fluentSS = 1.2MG∗20

∗8.34 = 200.2lbs

45036

lbs

MCRT: 6days =

lbs

day day

200.2

WASSS

lbs

day

45036

WASSS =

−200.2 = 7306

day

lbs

7306

7500∗8.34

7306

= WASFlow(MGD)∗7500∗8.34 =⇒ WASFlow(MGD) =

= 0.116MGD =

day

420

Chapter 22. Water Math

gal

day

116,000

45. Calculate the MCRT of an activated sludge plant given the following information.

Plant ﬂow- 4.25 MGD

WAS conc-7980 mg/l

Waste ﬂow- 0.055 MGD

RAS conc.- 7980 mg/l

Aeration tank vol-1MG

Clariﬁer vol- 0.25 MG

Final eff TSS conc. - 21.2 mg/l

MLSS conc.- 2050 mg/l

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 1∗2050∗8.34 = 17097lbs

MLSS in clarifier (lbs) = 0.25∗2050∗8.34 = 4274.3lbs

SS e f fluent (lbs/day) = 4.25MGD∗21.2mg/l ∗8.34 = 751.4lbs/day

SS WAS (lbs/day) = 0.055MGD∗7980mg/l ∗8.34 = 3660.4lbs/day

17097.6+4274.3

Plugging in the values calculated above: MCRT(days) =

= 4.8 = 5days

751.4+3660.4

22.27 Dewatering math problems

421

46. Calculate the MCRT given the following.

Plant ﬂow - 1.8 MGD

MLSS conc - 2800 mg/l

WAS ﬂow - 0.04 MGD

MLVSS conc. - 2190 mg/l

Aerator vol - 0.3 MG

Reactor vol. - 0.2 MG

RAS conc. - 8150 mg/l

Efﬂuent SS conc.-18 mg/l

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 0.3∗2800∗8.34 = 7005.6lbs

MLSS in clarifier (lbs) = 0.2∗2800∗8.34 = 4670.4lbs

SS e f fluent (lbs/day) = 1.8MGD∗18mg/l ∗8.34 = 270.2lbs/day

SS WAS (lbs/day) = 0.04MGD∗8150mg/l ∗8.34 = 2718.8lbs/day

7005.6+4670.4

270.2+2718.8

Plugging in the values calculated above: MCRT(days) =

= 3.9 = 4days

422

Chapter 22. Water Math

47. In an aeration tank, the MLSS is 2650 mg/l and recorded 30-minute settling test indicates 221 ml/L.

What is the sludge volume index?

Solution:

Settled sludge volume in ml/l after 30 min

MLSS mg/l

SVI (ml/g)=

∗1000

221ml/l

2650mg/l

SVI=

∗1000

= 83ml/g

48. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,

how many lbs of MLVSS should be maintained in the aeration tank?

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

2100

=⇒ 0.35 =

=⇒ x = 130,100lbs MLVSS

49. Calculate F/M ratio based on the following data:

Secondary inﬂuent BOD - 156 mg/l

Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep

Inﬂuent ﬂow - 0.65 MGD

MLSS - 3600 mg/l

MLSS average % volatile - 72%

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

156∗0.65∗8.34

F : M =

= 0.06

7.48gal

3600∗0.72∗4∗(30∗70∗10)ft³∗

∗

∗8.34

1000000gal

ft³

22.27 Dewatering math problems

423

50. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the

minimum MLSS concentration, given the following:

Q = 0.4 MGD

Primary inﬂuent BOD = 250 mg/l

Primary efﬂuent BOD = 128 mg/l

Aeration tank vol. = 350,000 gallons

Clariﬁer vol = 250,000 gallons

MLSS has 80% volatile solids

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

=⇒ F : M ∝

=⇒ F:M is inversely proportional to MLSS

MLSS concentration

So to have minimum MLSS conc. F:M needs to be the maximum of the range provided

0.4∗128∗8.34

If the MLSS concentration = x: =⇒ F : M = 0.28 =

=⇒ x = 5446mg/l MLSS

0.35∗x∗0.8

51. 5. Given that an activated sludge plant with an inﬂuent ﬂow of 1 MGD is operated at an MCRT of

6 days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day. (10

points)

Two aeration tanks – 0.45 MG each

Two ﬁnal clariﬁers – 0.2 MG each

WAS = 6500 ppm

Final efﬂuent = 16

MLSS volatile solids content = 85%

MLSS =3000

Solution:

lbsMLSS(system)

MCRT=

lbs

day

lbs

E f fluentSS+

WASSS

day

lbs MLSS (system)= (2∗0.45+2∗0.2)MG∗3000

∗8.34 = 32,526lbs

lbs

day

E f fluentSS = 1MG∗16

∗8.34 = 133.4lbs

32,526

lbs lbs

day day

MCRT: 6days =

133.4

WASSS

lbs

day

32,526

lbs

WASSS =

−133.4 = 5,288

day

lbs

day

5,288

= WASFlow(MGD)∗6500∗8.34

424

Chapter 22. Water Math

5,288

6,500∗8.34

gal

day

=⇒ WASFlow(MGD) =

= 0.097546MGD = 97,546

52. Operational data is given below for a conventional activated sludge treatment plant:

Inﬂuent ﬂow: 2.5 mgd

Inﬂuent BOD: 220 mg/L

Inﬂuent TSS: 240 mg/L

Primary BOD removal efﬁciency: 30%

Aeration tank volume: 1.8 MG

Secondary clariﬁer volume: 0.8 MG

MLSS: 3,600 mg/L

MLVSS: 2,800 mg/L

RAS SS: 8,500 mg/L

RAS VSS: 6,630 mg/L

RAS ﬂow: 100%

WAS ﬂow: 35,000 gpd

AS Efﬂuent TSS: 25 mg/L

AS Efﬂuent BOD: 19 mg/L

Settleability results: 60 min= 300 ml/L

Settleability results: 30 min= 320 ml/L

(a) Calculate the MCRT

(b) Calculate the F/M Ratio

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

(a) MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 1.8∗3,600∗8.34 = 54,043lbs

MLSS in clarifier (lbs) = 0.8∗3,600∗8.34 = 24,019lbs

SS e f fluent (lbs/day) = 2.5MGD∗25mg/l ∗8.34 = 521lbs/day

35,000

SS WAS (lbs/day) =

MGD∗8,500mg/l ∗8.34 = 2,481lbs/day

1,000,000

54,043+24,019

521+2,481

MCRT(days) =

= 26 days

(lbs/day) primary e f fluent BOD entering the aeration tank

(lbs) MLVSS in the aeration tank

(b) F : M =

F : M =

220∗(1−0.3)∗2.5∗8.34

= 0.08

1.8∗2,800∗8.34

Settled sludge volume in ml/l after 30 min

MLSS mg/l

∗1000

320ml/l

3,600mg/l

SVI=

∗1000

= 89ml/g

53. What is the Sludge Volume Index given the following:

MLSS = 2800 mg/L; MLVSS 2400 mg/l; Settled Volume after 30 minutes = 250 ml/L

22.27 Dewatering math problems

425

Solution:

Settled sludge volume in ml/l after 30 min

SVI (ml/g)=

∗1000

MLSS mg/l

250ml/l

2800mg/l

54. Plant Inﬂuent Flow: 3.5 MGD

SVI=

∗1000

= 89ml/g

Plant Inﬂuent BOD: 276 mg/l

Primary Treatment BOD Removal: 36%

Desired F/M: 0.3

Find lbs of MLVSS that should be maintained in the activated sludge treatment process. (Answer:

17,187 lbs)

55. Given that an activated sludge plant with an inﬂuent ﬂow of 1 MGD is operated at an MCRT of 6

days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day

Two aeration tanks – 0.45 MG each

Two ﬁnal clariﬁers – 0.2 MG each

WAS = 6500 ppm

Final efﬂuent = 16

MLSS volatile solids content = 85%

MLSS =3000

(Answer: 97,546 gal/day)

56. The aeration tank of a conventional activated sludge plant contains 9300 Ibs of MLSS. Lab tests

show that the MLSS are 79% volatile solids. The primary efﬂuent contains 2900 Ibs of BOD.

Calculate the F to M ratio for this wastewater treatment plant. (Answer: 0.4)

57. How many lbs of solids are in a 400,000 gallon aeration tank if the suspended solids concentration

is 1200 mg/l? (Answer: 4003 lbs)

58. An activated sludge wastewater treatment plant treats an average ﬂow of 3.25 MGD. The aeration

tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The ﬁnal clariﬁer has an

operating volume of 0.25 MG. Efﬂuent suspended solids concentration averages 21.2 mg/L. The

WAS concentration is 8100 mg/L and is wasted at the rate of 0.055 MGD. Calculate the MCRT.

59. What is the F/M ratio on an aeration tank if 1,500 pounds of BOD are added per day and 5000

pounds of volatile solids are under aeration?

60. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test indicates 230 ml.

What is the sludge volume index?

61. An activated sludge aeration tank receives a primary efﬂuent ﬂow of 2.61 MGD with a BOD con-

centration of 195 mg/L. The mixed liquor volatile suspended solids concentration is 2560 mg/L and

the aeration tank volume is 470,000 gallons. What is the current F/M ratio?

62. Calculate the MCRT given the following. Plant ﬂow- 2.8 MGD

MLSS conc.- 2800 mg/L

WAS ﬂow- 0.07 MGD

MLVSS conc.- 2190mg/L

Aerator volume - 0.38MG

RAS SS concentration - 6800 mg/L

Clariﬁer volume - 0.19 MG

Final efﬂuent TSS - 18 mg/L

63. If in a conventional activated sludge treatment plant the aeration tank contains 7000 lbs of MLSS

and the ﬁnal clariﬁer contains 2500 lbs of MLSS. If 1300 lbs of solids are wasted each day and 120

426

Chapter 22. Water Math

lbs of solids leave in the ﬁnal efﬂuent, Calculate the MCRT.

64. The aeration tank of a conventional activated sludge plant contains 9300 Ibs of MLSS. Lab tests

show that the MLSS are 79% volatile solids. The primary efﬂuent contains twenty-nine hundred

pounds (2900 Ibs) of BOD. Calculate the F to M ratio for this wastewater treatment plant.

65. A conventional activated sludge plant receives an average ﬂow of 2.5 MGD. The inﬂuent BOD

averages 230 and the primary efﬂuent BOD average 160 mg/I. The 0.6 MG aeration tank has a

MLSS cone. of 2800 mg/L and a MLVSS cone. of 21 50 mg/L. Calculated the F to M ratio for this

plant.

66. A convention activated sludge plant receives an average inﬂuent ﬂow of 800,000 gpd. The inﬂuent

BOD averages 200 mg/L and on the average 32% of this BOD is removed in the primary sedimen-

tation tank. The mixed liquor tank contains 3250 Ibs of MLSS. These solids contain 77% volatile

matter. Calculate this plant’s F to M ratio.

67. In an conventional activated sludge plant calculations show that the aeration tank contains 6500 Ibs

of MLSS and the ﬁnal clariﬁer contains 2500 Ibs of MLSS. Twelve hundred and ﬁfty pounds (1250

Ibs) of solids is wasted each day and 100 Ibs/day of solids leave in the ﬁnal efﬂuent. Calculate the

MCRT for this plant.

68. A activated sludge plant has a total of 25,000 pounds of solids in the system (i.e. aerator + ﬁnal

clariﬁer) and has an average efﬂuent ﬂow of 3.75 MGD. The efﬂuent suspended solids concentra-

tion averages 22 mg/L. The WAS has a concentration of 7500 mg/L and is being wasted at 0.05

MGD. Calculate the MCRT.

69. An activated sludge wastewater treatment plant discharges an average ﬂow of 3.25· MGD. The

mixed liquor tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The ﬁnal clar-

iﬁer has an operating volume of 0.25 MG. Efﬂuent suspended solids concentration averages 21.2

mg/L. The WAS concentration is 8100 j mg/L and is wasted at the rate of 0.055 MGD. Calculate

the MCRT.

70. In an conventional activated sludge plant show that the aeration tank contains 6500 Ibs of MLSS

and the ﬁnal clariﬁer contains 2500 lbs of MLSS. 1250 Ibs of solids is wasted each day and 100

Ibs/day of solids leave in the ﬁnal efﬂuent. Calculate the MCRT for this plant. Answer: 7 days

71. An activated sludge plant has a total of 25,000 pounds of solids in the system (aerator + ﬁnal Clar-

iﬁer) and has an average efﬂuent ﬂow of 3.75 MGD. The efﬂuent suspended solids concentration

averages 22 mg/L. The WAS has a concentration ’of 7500 .mg/L and is being wasted at 0.05 MGD.

Calculate the MCRT. Answer: 7 days

72. An activated sludge wastewater treatment plant discharges an average ﬂow of 3.25 MGD. The

mixed liquor tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The ﬁnal clar-

iﬁer has an operating volume of 0.25 MG. Efﬂuent suspended solids concentration averages 21.2

mg/L and the WAS concentration is 8100 mg/L and is wasted at the rate of 0.055 MGD. Calculate

the MCRT. Answer: 5.1 days

73. A conventional activated sludge plant receives an average ﬂow of 2.5 MGD. The inﬂuent BOD

averages 230 and the primary efﬂuent BOD averages 160mg/L The 0.6 MG aeration tank has a

MLSS conc. of 2800 mg/L and a MLVSS conc. of 2150mg/L. Calculate the F to M ratio for this

plant. Answer: 0.31

74. A conventional activated sludge plant receives an inﬂuent ﬂow of 800,000 gpd. The inﬂuent BOD

averages 200 mg/L and on the average 32% of this BOD is removed in the primary sedimentation

tank. The mixed liquor tank contains 3250 Ibs of MLSS. These solids contain 77% volatile matter.

Calculate this plant’s F to M ratio. Answer: 0.36

22.27 Dewatering math problems

427

75. Calculate the MCRT given the following. Plant ﬂow- 1.8 MGD

MLSS conc.- 2800 mg/L

WAS ﬂow- 0.04 MGD

MLVSS conc.- 2190mg/L

Aerator vol- 0.3MG

RAS conc.- 8150 mg/L

Clariﬁer vol- 0.lMG

Final eff TSS- 26 mg/L

Answer: 3 days

76. Calculate the F to M ratio given the following Information. Plant ﬂow- 3.7 MGD

Primary eff TSS- 120mg/L

Aerator Vol - 0.85 MG

MLSS conc.- 3200 mg/L

Clariﬁer Vol - 0.25 MG

MLVSS conc.- 2550 mg/L

Primary efﬂuent BOD-120 mg/L

Answer: 0.2

77. Calculate the F to M ratio given the following. Plant ﬂow- 0.8 MGD Aerator vol- 200,000 gal.

Clariﬁer Volume-175,000 gal Primary eff BOD- 120 mg/L MLVSS cone. -1950 mg/L

Answer: 0.25

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

120∗0.8∗8.34

F : M =

= 0.25

1950∗0.2∗8.34

78. Calculate the MCRT of an activated sludge plant given the following information. Plant ﬂow- 4.25

MGD

WAS conc-7980 mg/L

Waste ﬂow- 0.055 MGD

RAS conc.- 7980 mg/L

Aeration tank vol-1MG

Clariﬁer vol- 0.25 MG

Final eff TSS conc. - 21.2 mg/L

MLSS conc.- 2050 mg/L

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 1∗2050∗8.34 = 17097lbs

MLSS in clarifier (lbs) = 0.25∗2050∗8.34 = 4274.3lbs

SS e f fluent (lbs/day) = 4.25MGD∗21.2mg/l ∗8.34 = 751.4lbs/day

SS WAS (lbs/day) = 0.055MGD∗7980mg/l ∗8.34 = 3660.4lbs/day

17097.6+4274.3

Plugging in the values calculated above: MCRT(days) =

= 4.8 = 5days

751.4+3660.4

79. Calculate the MCRT given the following.

Plant ﬂow - 1.8 MGD

428

Chapter 22. Water Math

MLSS conc - 2800 mg/L

WAS ﬂow - 0.04 MGD

MLVSS conc. - 2190 mg/L

Aerator vol - 0.3 MG

Reactor vol. - 0.2 MG

RAS conc. - 8150 mg/L

Efﬂuent SS conc.-18 mg/L

Solution:

MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)

MCRT(days) =

SS e f fluent (lbs/day)+SS WAS (lbs/day)

MLSS in aeration tank (lbs) = 0.3∗2800∗8.34 = 7005.6lbs

MLSS in clarifier (lbs) = 0.2∗2800∗8.34 = 4670.4lbs

SS e f fluent (lbs/day) = 1.8MGD∗18mg/l ∗8.34 = 270.2lbs/day

SS WAS (lbs/day) = 0.04MGD∗8150mg/l ∗8.34 = 2718.8lbs/day

7005.6+4670.4

270.2+2718.8

Plugging in the values calculated above: MCRT(days) =

= 3.9 = 4days

80. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,

how many lbs of MLVSS should be maintained in the aeration tank?

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

2100

=⇒ 0.35 =

=⇒ x = 6000lbs MLVSS

81. Calculate F/M ratio based on the following data:

Secondary inﬂuent BOD - 156 mg/l

Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep

Inﬂuent ﬂow - 0.65 MGD

MLSS - 3600 mg/l

MLSS average % volatile - 72%

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

156∗0.65∗8.34

F : M =

= 0.06

7.48gal

3600∗0.72∗4∗(30∗70∗10)ft³∗

∗

∗8.34

1000000gal

ft³

82. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the

minimum MLSS concentration, given the following: Q = 0.4 MGD Primary inﬂuent BOD = 250

mg/L Primary efﬂuent BOD = 128 mg/L Aeration tank vol. = 350,000 gallons Clariﬁer vol =

250,000 gallons MLSS has 80% volatile solids Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

=⇒ F : M ∝

MLSS concentration

22.27 Dewatering math problems

429

So to have minimum MLSS conc. F:M needs to be the maximum of the range provided

0.4∗128∗8.34

If the MLSS concentration = x: =⇒ F : M = 0.28 =

=⇒ x = 5446mg/l MLSS

0.35∗x∗0.8

83. If the target MCRT for an activated sludge plant with the following parameters is 8 days, what is

the required wasting rate (in gallons per day)? (8 points)

Aeration Tank Volume = 1.5 MG

Clariﬁer Volume = 1.2 MG

MLSS Concentration = 2500 mg/l

MLSS Volatile Solids = 79%

WAS SS = 5100 mg/l

Secondary Efﬂuent SS = 12 mg/l

Plant ﬂow = 15 MGD

(Total MLSS lbs in the aeration system(aeration tank + clari fier))

MCRT =

lbs

day

(Total amount in

of SS leaving the system(e f fluent +WAS))

(MLSS_{aeration tank}(lbs) + MLSS_clariﬁer(lbs))

(SS_efﬂuent(lbs/day)+SS_WAS(lbs/day))

MCRT =

(1.5+1.2)MG∗8.34∗2500

=⇒ 8days =

(15MGD∗12

∗8.34)(lbs/day SS_efﬂuent+(xMGD WAS∗5100

∗8.34)(lbs/day SS_WAS

)

(1.5+1.2)MG∗8.34∗2500

(15∗12∗8.34(lbs/day)SS_efﬂuent+xMGD WAS∗5100∗8.34)(lbs/day)SS_WAS

(1.5+1.2)MG∗8.34∗2500

xMGD WAS∗5100∗8.34(lbs/day)SS_WAS

5535.7

−(15∗12∗8.34(lbs/day)SS_efﬂuent

gal

42,534x = 5535.7 =⇒ x =

= 0.1301MGD = 130,100

42,534

day

84. Given the following: Plant Inﬂuent Flow: 3.5 MGD

Plant Inﬂuent BOD: 276 mg/l

Primary Treatment BOD Removal: 36%

Desired F/M: 0.3

Find lbs of MLVSS that should be maintained in the activated sludge treatment process. (7 points) F :

lbs In fluent BOD

lbs MLVSS

276∗(1−0.36)∗3.5∗8.34

M =

=⇒ 0.3 =

=⇒ lbs MLVSS = 17,187lbs

lbsMLVSS

5. Given that an activated sludge plant with an inﬂuent ﬂow of 1 MGD is operated at an MCRT of

6 days and the parameters below, calculate the WAS ﬂow rate (wasting rate) in gallon per day. (10

points)

Two aeration tanks – 0.45 MG each

Two ﬁnal clariﬁers – 0.2 MG each

WAS = 6500 ppm

Final efﬂuent = 16

MLSS volatile solids content = 85%

MLSS =3000

Solution:

lbsMLSS(system)

MCRT=

lbs

day

lbs

E f fluentSS+

WASSS

day

lbs MLSS (system)= (2∗0.45+2∗0.2)MG∗3000

∗8.34 = 32,526lbs

430

Chapter 22. Water Math

lbs

day

E f fluentSS = 1MG∗16

∗8.34 = 133.4lbs

32,526

lbs lbs

day day

MCRT: 6days =

133.4

WASSS

lbs

day

32,526

lbs

WASSS =

−133.4 = 5,288

day

lbs

day

5,288

= WASFlow(MGD)∗6500∗8.34

5,288

gal

day

=⇒ WASFlow(MGD) =

= 0.097546MGD = 97,546

6,500∗8.34

85. The desired F/M ratio is 0.35. If 2,100 lbs of BOD enter the aerator daily and the volume of the

mixed liquor in the aeration tank is 375,000 gallons , what should be the target MLVSS concentra-

tion?

Solution:

(lbs/day) primary e f fluent BOD entering the aeration tank

F : M =

(lbs) MLVSS in the aeration tank

2100

=⇒ 0.35 =

=⇒ x = 6000lbs MLVSS

=⇒ 6,000lbs MLVSS = 8.34∗ _1,000,000MG∗x^mg

375,000

MLVSS

mg MLVSS

=⇒ x^mg

= 1,900

MLVSS

6000

8.34∗0.375

Solids Treatment Math Problems

1. Calculate the volatile solids reduction in an anaerobic digester given the following information:

Raw sludge feed to digester: 73.7 % VS and digested sludge: 57.2 %VS

*a. 52.3%

b. 64%

c. 16.5%

d. 42%

Solution:

The VS reduction of the digester is provided by the Van Kleeck equation

VS_in−VS_out

Digester VS reduction(%) =

∗100

VS_in−VS_in∗VS_out

0.737−0.572

Digester VS reduction(%) =

∗100 = 52.3%

0.737−0.737∗0.572

2. 42,000 gallons of 6% sludge containing 67% volatile matter is pumped to the digester. The di-

22.27 Dewatering math problems

431

gester reduces the volatile matter by 52%. What volume of sludge in gallons containing 5% solids

remains after digestion? [Ans: 32,841 gal]

3. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge ﬂow. The

sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic

foot per day? [0.057 lbs VS/ft3-day]

4. An Imhoff cone result is 5.5 ml/l. Approximately how many gallons of primary sludge will need to

be pumped if the ﬂow is 0.7 MGD? [3,850 gallons]

5. A sludge digester equipped with a ﬂoating cover is 31 ft. inside diameter. The corbels are 16 feet

above the ﬂoor and the top of the wall is 20 feet above the ﬂoor. The digester currently has a sludge

depth of 16.5 ft. How many gallons of sludge are needed to displace the cover 1 foot? [5,643

galllons]

6. 10,000 gallons/day of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). If

50% of the VS is destroyed, how many lbs of VS is destroyed per day?

Solution:

1,168lbs VS destroyed

10,000 Gal 8.34 lbs sludge 0.04∗0.7 lbs VS feed 0.5 lbs VS destroyed

∗

day

Gal

lb sludge

lbs VS feed

day

7. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.

secondary sludge at 1.3% solids. What is the combined solids concentration? [Ans. 2.28%] If the

primary sludge is 68% VS and the secondary sludge is 63% VS, how many pounds of VS are in the

combined sludge? [Ans. 5400.3 lbs VS]

8. 12,000 gallons of 1.8% sludge is pumped to a thickener, and is thickened to 3800 gallons of 4.9%

solids. The supernatant is returned back to the head of the STP for treatment. What is the volume in

gallons and solids concentration in ppm of the supernatant?[Ans. 24,980 gallons]

9. How many pounds of solids are pumped to a digester each day if the digester receives 10,000 gpd

of sludge at 5% solids concentration?

Solution:

lbs TS 10,000gal sludge (8.34∗0.05lbsTS)

= 4,170

lbs TS

day

∗

day

gal sludge

10. Calculate the % VS reduction in a digester given the volatile solids content of the inﬂuent sludge to

the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%

0.7−0.525

Solution: Digester VS reduction(%) =

∗100 = 53%

0.7−0.7∗0.525

11. If an anaerobic digester receives sludge with VS of 80% and discharges digested sludge with a 60%

VS. Its VS reduction is:

a. 20%

b. 25%

c. 63%

d. 89%

Solution:

0.8−0.6

Digester VS reduction(%) =

∗100 = 63%

0.8−0.8∗0.6

12. Calculate the VS loading to the digester in lbs/day if 10,000 gallons of sludge containing 5% TS

with and average VS content of 78%

Solution:

Digester VS loading (lbs/day)

432

Chapter 22. Water Math

10,000 gallons sludge 8.34lbs sludge 0.05∗0.78lbsVS

∗

= 3,253lbs sludge per day

day

gal

lb sludge

13. Primary sludge containing ﬁve percent (5%) solids is pumped to a digester continuously at a rate

of 25 gpm. How many pounds of volatile solids are added to the digester each day if the volatile

solids content is 73% of the total solids?

a. 1,310 lbs/day

b. 1,800 lbs/day

c. 9,830 lbs/day

d. 10,960 lbs/day

e. 15,010 lbs/day

Solution:

Digester VS loading (lbs/day)

25∗1,440 gallons sludge 8.34lbs sludge 0.05∗0.73lbsVS

∗

= 10,960 lbs sludge per day

day

gal

lb sludge

14. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge ﬂow. The

sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic

foot per day?

Solution:

lbs VS

Digester Loading

day

Digester volatile solids loading rate =

Digester volume(V) ft³

gal sludge

day

lbsVS

gal sludge

5000

∗(8.34∗0.06∗0.66)

lbs VS

day− ft³

= 0.057

(

∗37²∗27)ft³

15. How much gas is produced in the above digester in ft³/day if the digested sludge contains 2.5%

total solids of which 59% is volatile solids and the gas production rate is 14 ft³/lb VS destroyed?

Solution:

0.70−.59

Digester VS reduction(%) =

∗100 = 38.3%

0.70−0.70∗0.59

lbs VS reduction 3153lbs VS feed ∗0.383 VS reduction

lbs VS reduction

day

= 1,208

day

ft³gas produced

day

lbs VS reduced 14 ft³gas produced

ft³digester gas produced

= 1208

∗

= 16,912

day

lb VS reduced

day

16. The sludge feed to a digester is 80,000 gal/day. The sludge contains 4.5% total solids with 75%

volatile solids. If 50% of VS are reduced in the digester:

(a) Find lbs VS destroyed per 1000 gal of digester capacity per day if the digester radius is 55 ft

with an operating sludge level of 25 ft.(5 points)

22.27 Dewatering math problems

433

Solution:

gal

ft³

Digester Volume: (π ∗55²∗25) ft³∗7.48

= 1,776,220gallons = 1,776.2 (1000 gallons)

lbs

gal

lbsVS

lbs VS reduction

∗0.5

lb VS

80,000gal ∗8.34

∗(0.045∗0.75)

lbs VS reduction

day

lbs VS reduction

day

= 11,259

lbs VS reduction

day

1,776.2 (1000 gallons)

11,259

lbs VS reduction

1000 digester capacity

lbs VS reduction

1000gallons digester volume

= 6.3

(b) What is the digester gas production in Btu/day? Assume 14 cu. ft digester gas per lb of VS

destroyed and a 650 Btu/cu. ft heating value for the digester gas produced (5 points)

Solution:

ft³gas produced

day

lbs VS reduced 14 ft³gas produced

ft³digester gas produced

day

= 11,259

∗

= 157,626

day

lb VS reduced

BTU produced

ft³gas produced 650BTU gas produced

BTU produced

= 157,626

∗

= 102,456,900

day

ft³gas

day

17. The volatile acids concentration of sludge in an anaerobic digester is 195 mg/l. If the maximum

volatile acids:alkalinity ratio is 0.087, what should the alkalinity be in mg/l?

Solution:

195

0.087

Volatile acid:Alkalinity=0.087=

=⇒ x =

= 2,241

434

Chapter 22. Water Math

18. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of

the VS are destroyed, creating 15 ft³of gas per lbs. of VS destroyed. How much gas is produced

each day?

lbs

gal

lbsVS 0.5lbs VS reduction

10,000gal ∗8.34

∗(0.04∗0.7)

∗

lbs VS reduction

day

lbs VS reduction

day

lb VS

= 1,168

day

ft³gas produced

day

lbs VS reduced 15 ft³gas produced

ft³digester gas produced

day

= 1,168

∗

= 17,520

day

lb VS reduced

19. Given a sludge ﬂow of 5,500 gpd with 6.5% solids containing 72% VS to a digester with a VS

destruction of 48% If the digester is digester is 40 ft. diameter with a sludge level of 25 ft, ﬁnd

lbs. VS destroyed per 1000 gal of digester capacity per day. What is the digester gas production in

Btu/day if the % VS reduction in the digester is 58% and the digester raw sludge feed is 6000 cu.

Ft/day containing 4.5% total solids with a 64% VS content? Assume 13.5 cu. ft digester gas per lb

of VS destroyed and a 650 Btu/cu. ft heating value for the digester gas produced

gal

ft³

Digester Volume: ( ∗40²∗25)ft³∗7.48

= 940,086gallons = 940 (1000 gallons)

lbs

gal

lbsVS 0.48lbs VS reduction

5,500gal ∗8.34

∗(0.065∗0.72)

∗

lbs VS reduction

day

lb VS

day

lbs VS reduction

day

1,030

lbs VS reduction

day

940 (1000 gallons)

1,030

lbs VS reduction

1000 digester capacity

lbs VS reduction

1000gallons digester volume

= 1.1

ft³gas produced

lbs VS reduced 13.5 ft³gas produced

ft³digester gas produced

day

= 1,030

∗

= 13,905

day

lb VS reduced

BTU produced

day

ft³gas produced 650BTU gas produced

BTU produced

= 13,905

∗

= 9,038,250

day

ft³gas

day

20. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and

28,000 gal secondary sludge at 5% solids.

(a) What is the combined solids concentration? (3 points)

60,000∗4.5+28,000∗5

= 4.66%

88,000

(b) How many lbs of solids (per day) are in the combined sludge (3 points)

88,000gal ∗8.34∗0.0466 = 34,200lbs

VS are in the combined sludge? (3 points)

22.27 Dewatering math problems

8.34∗0.045∗0.68lbs VS

435

8.34∗0.05∗0.83lbs VS

60,000gal∗

+28,000gal∗

= 25,003lbs VS

gal

(d) What is the digester feed VS%? (3 points)

25,003lbs VS

34,200lbsTS

= 73.1%

(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs

VS/ft3-day (3 points)

25,003lbs VS

0.785∗120²∗20

0.11lbs VS

day− ft³

(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)

0.731−.52

Digester VS reduction(%) =

∗100 = 60.1%

0.731−0.731∗0.52

(g) What is the gas production (ft³/day) if the digester produces 14 ft³gas/lb VS destroyed (3

points)

lbs VSdestroyed 14 ft³gas produced

25,003*0.601

∗

= 210,375ft³gas

lb VS destroyed

(h) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

can be operated at a maximum ﬂow of 50 GPM (3 points)

gal sludge

day

1440min

88,000

∗

= 61GPM

@50 GPM per press - 2 BFP required

(i) If the digested sludge feed to the belt ﬁlter presses is 2.6% and assuming the belt press feed

solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)

gal

day

lbs solids feed

day

lbs solids captured

lbs solids feed

88,000

∗8.34∗0.026

∗0.9

= 17,174lbs solids per day

436

Chapter 22. Water Math

(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)

17,174lbs solids produced

100lbs cake

20lbs solids produced 2,000lbs

tons

42.9tons

day

∗

day

(k) If the cake density is 68 lbs/cu. ft, how much time will it take to ﬁll a 30 cu. yd bin (3 points)

(42.9∗2000)lbs cake

day

59.583

(Ans. 15.4 hours)

∗

min

day

1440min lbs cake

59.583

min∗

ft³

0.876 ft³

lbs cake

68lbs cake

min

27 ft³

yd³

min

hrs

30yd³∗

∗

0.876ft³60min

= 15.4hrs

(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)

tons $65

day ton

$2,789

day

42.9

∗

21. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS

with and average VS content of 76% is fed to the digester

22. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge

feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge

feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-

ft3 )

23. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given:

Primary sludge feed rate of 25,000 gallons with 3.8% TS containing 73% solids and SG of 1.03.

Secondary sludge feed rate of 30,500 gallons with 3.8% TS containing 77% solids

lbs VS

day− ft³

(Answer: 0.2

)

24. 325 GPM of WAS with 7,000 mg/l TSS is thickened in a DAFT. Assuming an air:solids ratio of

0.05 lb of air for each lb of feed solids, The SCFM (standard cubic feed per minute) air required to

thicken the WAS given 1 SCFM = 0.075 lb is:

(Answer: 12.6 SCFM)

Solution:

Correct Answer:

0.075 lbs air

∗x SCFM

lb air

lb solids

SCF

0.05 =

325

1,000,000

(

MG per min∗7000∗8.34) lbs solids

0.05∗325∗7000∗8.34

0.075∗1,000,000

=⇒ x SCFM =

= 12.6 SCFM −Correct answer

Incorrect Answer #1:

12,600SCFM

22.27 Dewatering math problems

437

Incorrect Answer #2:

0.126SCFM

Incorrect Answer #3:

1.26SCFM

25. 30,000 gpd of 4.5% sludge is dewatered in a centrifuge yielding 24.5 yd³/day of 25% cake with a

density of 65 lb per ft³. Calculate the percentage solids recovery.

(Answer: 95.5%)

BTU

day

26. Estimate the amount of heat value of the gas produced from an anaerobic digester in

the following:

given

Raw sludge pumping schedule

Sludge pumping rate

Raw sludge %TS

12 min/hr

68 GPM

5.2%

Raw sludge %VS

Gas production

72.5%

12 ft³

lb VS destroyed − day

Percent CO₂

34%

Other gases

BTU

Pure methane net heat value

932

ft³

BTU

day

(Answer: 23,146,220

)

27. 12,000 ft³of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The

centrifuge yields 37 yd³of 26% of dewatered cake with a density of 73 lb/ft³. Calculate the solids

capture rate.

Solution:

gal (8.34∗0.028lbs TS)

= 20,960lbs TS

lbs TS feed to centrifuge = 12,000ft³sludge∗7.48

∗

ft³

gal sludge

ft³(73lbs∗0.26 TS)

lbs TS feed from centrifuge = 37yd³sludge∗27

∗

= 18,961lbs TS

yd³

ft³sludge

18,961lbs solids produced by centri fuge

solids capture rate =

∗100 = 90.4%solids capture

20,960lbs solids fed from digester

28. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at

68 lbs/ ft³. How long would it take to ﬁll a 3 yd³bin

Solution:

Approach:

• First calculate the amount of cake solids produced in terms of weight per time.

• From the weight of the cake produced calculate the volume time, and ﬁnally

•

using the calculated value of the volume of the cake produced per time, calculate the time

required to ﬁll the bin

cake TS produced −lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS

∗

∗0.9

lb sludge

min

galllon

12.61lbs TS

min

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

438

Chapter 22. Water Math

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

0.927 ft³

27 ft³

yd³

Time required to fill the bin =

∗3yd³∗

= 75min

22.27 Dewatering math problems

439

29. Calculate the annual cost savings for an improvement of cake solids from 20% to 21% if 400 wet

tons (@ 20% cake) are produced each day and the average cake hauling and disposal cost is $65

per wet ton.

400 wet tons 20tons solids

Total (dry) solids produced per day =

∗

= 80tons solids

day

100wet tons

80 tons solids 100wet tons

Tons wet cake @21% solids =

∗

day

21tons solids

380tons wet solids

day

wet tons

day

days

$65

wet ton

Net savings ($/yr) = (400−380)

∗365∗

∗

= $474,500 per year

NOTE: General formula for future reference: So if the cake dryness goes up from 20% to

26% and currently a utility is spending $1,000,000 per year for biosolids hauling and disposal,

their net savings will be: (26-20)/26*1,000,000 = $230,769. Conversely say if the dewatering

cake solids drops to 18%, the net impact will be: (18-20)/18*1,000,000 = $-111,111 (loss or

extra cost).

30. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and

28,000 gal secondary sludge at 5% solids.

(a) What is the combined solids concentration? (3 points)

60,000∗4.5+28,000∗5

= 4.66%

88,000

(b) How many lbs of solids (per day) are in the combined sludge (3 points)

88,000gal ∗8.34∗0.0466 = 34,200lbs

VS are in the combined sludge? (3 points)

8.34∗0.045∗0.68lbs VS

8.34∗0.05∗0.83lbs VS

60,000gal∗

+28,000gal∗

= 25,003lbs VS

gal

(d) What is the digester feed VS%? (3 points)

25,003lbs VS

34,200lbsTS

= 73.1%

(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs

VS/ft3-day (3 points)

25,003lbs VS

0.785∗120²∗20

0.11lbs VS

day− ft³

(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)

0.731−.52

Digester VS reduction(%) =

∗100 = 60.1%

0.731−0.731∗0.52

(g) What is the gas production (ft³/day) if the digester produces 14 ft³gas/lb VS destroyed (3

points)

440

Chapter 22. Water Math

= 210,375ft³gas

lbs VSdestroyed 14 ft³gas produced

25,003*0.601

∗

lb VS destroyed

(h) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

can be operated at a maximum ﬂow of 50 GPM (3 points)

gal sludge

day

1440min

88,000

∗

= 61GPM

@50 GPM per press - 2 BFP required

(i) If the digested sludge feed to the belt ﬁlter presses is 2.6% and assuming the belt press feed

solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)

gal

day

lbs solids feed

day

lbs solids captured

lbs solids feed

88,000

∗8.34∗0.026

∗0.9

= 17,174lbs solids per day

(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)

17,174lbs solids produced

100lbs cake

20lbs solids produced 2,000lbs

tons

42.9tons

day

∗

day

(k) If the cake density is 68 lbs/cu. ft, how much time will it take to ﬁll a 30 cu. yd bin (3 points)

(42.9∗2000)lbs cake

day

59.583

(Ans. 15.4 hours)

∗

min

day

1440min lbs cake

59.583

min∗

ft³

0.876 ft³

lbs cake

68lbs cake

min

27 ft³

yd³

min

hrs

30yd³∗

∗

0.876ft³60min

= 15.4hrs

(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)

tons $65

day ton

$2,789

day

42.9

∗

31. 12,000 ft³of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The

centrifuge yields 37 yd³of 26% of dewatered cake with a density of 73 lb/ft³. Calculate the solids

capture rate.

Solution:

gal (8.34∗0.028lbs TS)

= 20,976lbs TS

lbs TS feed to centrifuge = 12,000ft³sludge∗7.48

∗

ft³

gal sludge

ft³(73lbs∗0.26 TS)

lbs TS feed from centrifuge = 37yd³sludge∗27

∗

= 18,961lbs TS

yd³

ft³sludge

18,961lbs solids produced by centri fuge

solids capture rate =

∗100 = 90.4%solids capture

20,976lbs solids fed from digester

32. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at

68 lbs/ ft³. How long would it take to ﬁll a 3 yd³bin

Solution:

Approach:

• First calculate the amount of cake solids produced in terms of weight per time.

• From the weight of the cake produced calculate the volume time, and ﬁnally

•

using the calculated value of the volume of the cake produced per time, calculate the time

required to ﬁll the bin

cake TS produced −lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS

∗

∗0.9

lb sludge

min

galllon

22.27 Dewatering math problems

441

12.61lbs TS

min

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

ft³cake produced

min

12.61lbs TS 100lbs cake

ft³cake

68lbs cake

0.927 ft³cake

∗

min

20lbs TS

min

0.927 ft³

27 ft³

yd³

Time required to fill the bin =

∗3yd³∗

= 75min

442

Chapter 22. Water Math

33. Calculate the annual cost savings for an improvement of cake solids from 20% to 21% if 400 wet

tons (@ 20% cake) are produced each day and the average cake hauling and disposal cost is $65

per wet ton.

Solution:

400 wet tons 20 tons solids

Total (dry) solids produced per day =

∗

= 80 tons solids

day

100 wet tons

80 tons solids 100 wet tons

Tons wet cake @21% solids =

∗

day

21 tons solids

380 tons wet solids

day

wet tons

day

days

$65

wet ton

Net savings ($/yr) = (400−380)

∗365∗

∗

= $474,500 per year

NOTE: General formula for future reference: So if the cake dryness goes up from 20% to

26% and currently a utility is spending $1,000,000 per year for biosolids hauling and disposal,

(26−20)

their net savings will be:

∗$1,000,000 = $300,000

34. A belt press is used for dewatering 70 GPM digested sludge containing 3% TS, seven hours per day.

At the end of the day it produces 24 yd³of 16% TS biosolids @ 65 lbs/ft³density. What is the

percent belt press solids recovery?

Solution:

70 gal 8.34 lbs 0.03 lbs solids 7∗60 min 7,356 lbs solids

lbs solids feed to belt press:

∗

min

gal

lb feed sludge

day

24 yd³cake 27 ft³65 lbs cake 0.16 lbs solids 6,739 lbs solids

lbs solids in belt press cake:

∗

day

yd³

ft³

lb cake

day

6,739

7,356

belt press solids recovery:

= 0.92 or 92%

35. Calculate the air required (SCFM) to meet a 0.04 lb air:lb feed solids ratio for a 100 GPM WAS

ﬂow with a solids content of 6500mg/l? Assume 0.08 lbs air/SCF air.

Solution:

0.08 lbs air

∗x SCF per minute

lb air

SCF

0.04 =

100

lb solids

(

MG per min∗6500mg/l ∗8.34) lbs solids

1,000,000

0.04

0.0148

=⇒ 0.04 = 0.0148 ∗ x SCF per minute =⇒ x SCF per minute =

= 2.7 SCFM

36. A treatment plant receives an inﬂuent ﬂow of 30 MGD with a TSS concentration of 280 mg/l. The

primary treatment removes 55% TS and the primary sludge is pumped to a 40 ft diameter gravity

thickener. Calculate the average solids loading to the thickener in lbs TSS/day-ft²

Solution:

(30 MGD ∗ 280∗0.55 mg/l ∗8.34)lbs TSS per day

Solids loading to gravity thickener=

0.785∗40²ft²

30.7 lbsTSS/day− ft²

22.27 Dewatering math problems

443

37. Two, 110’ diameter digesters with a cone depth of 15 ft and operating at a cylindrical liquid level of

28 ft is fed with a blend of primary and secondary sludge. The 85,000 gal/day primary sludge feed

contains 5% solids with a 65% VS content. The secondary sludge ﬂow is 33,000 gallons per day

with a 5.5% solids and 86% VS content.

(a) What is the combined solids concentration?

Combined solids concentration:

C₁∗V₁+C₂∗V₂= C₃∗V₃

C₁∗V₁+C₂∗V₂C₁∗V₁+C₂∗V₂5∗85,000+5.5∗33,000

= 5.14% TS content

=⇒ C₃=

V₁+V₂

88,000+33,000

(b) What is the digester VS loading rate (lbs VS/ ft³)

lbs VS

Digester VS loading =

Digester volume

8.34∗(Pri. Sldg gal/day∗Pri.Sldg TS∗%VS+ Sec. sludge gal/day∗Sec. sludgeTS∗%VS)

=⇒

3.14∗D²∗h₁

2∗

+0.785∗D²∗h

8.34∗(88,000 gal/day∗0.05∗0.65 + 33,000 gal/day∗0.055∗0.86)

lbs VS

ft³

= 0.059

(3.14∗110²∗15

2∗

+0.785∗110²∗28

lbs VS

lbs TS

Calculate VS% in: =

∗100

8.34∗(Pri. Sldg gal/day∗Pri.Sldg TS∗%VS+ Sec. sludge gal/day∗Sec. sludgeTS∗%VS)

8.34∗(Pri. Sldg gal/day∗Pri.Sldg TS+ Sec. sludge gal/day∗Sec. sludgeTS∗)

=⇒

8.34∗(88,000 gal/day∗0.05∗0.65 + 33,000 gal/day∗0.055∗0.86)

∗100

8.34∗(88,000 gal/day∗0.05 + 33,000 gal/day∗0.055)

36,870lbs VS

_{51,833 lbs TS}∗100 = 71.1%VS_in

0.71−.52

Digester VS reduction(%) =

∗100 = 55.8% VS reduction

0.71−0.71∗0.52

(d) What is the gas production (ft³/day) if the digester produces 14 ft³gas/lb VS destroyed

14 ft³gas produced

lb VS destroyed

36,870 lbs VS_in∗0.558lbs VS destroyed∗

= 288,028 ft³gas produced

(e)

(f) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

can be operated at a maximum ﬂow of 50 GPM

gal sludge

day

1440min

121,000

∗

= 84GPM

@50 GPM per press - 2 BFP required

(g) If the digested sludge feed to the belt ﬁlter presses is 2.6% and assuming the belt press feed

solids capture is 90%. How many lbs of solids (dry) are produced by the BFP

gal

day

lbs solids feed

day

lbs solids captured

lbs solids feed

121,000

∗8.34∗0.026

∗0.9

= 23,028lbs solids per day

(h) If the belt press produces a 20% cake at 68 lbs/ ft³. How long would it take to ﬁll a 3 yd³bin?

444

Chapter 22. Water Math

lbs cake

lbs solids

day

lb cake

0.2 lb solids

lbs cake : 23,028

∗

= 115,140

day

lbs cake

ft³

yd³

day

yd³cake

Volume of cake produced: 115,140

∗

= 2.61

day

68 lbs cake 27 ft³24 hrs

= 1.15 hrs

(i) Time to ﬁll the 3 cu. yd bin: 3 yd³∗

2.61yd³cake

eak

38. Two sludges are blended together as follows: 60,000 gal. primary sludge at 4.5% solids and 28,000

gal. secondary sludge at 5% solids.

(a) What is the combined solids concentration?

(b) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of

VS are in the combined sludge?

lbs VS/ft³-day?

(d) If the digester VS reduction is 60%, what is the gas production (ft³/day) if the digester

produces 14 ft³gas/lb VS destroyed?

(e) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

are operated at a maximum ﬂow of 50 GPM ?

(f) Knowing the digester VS reduction and the TS feed, calculate the solids (TS) concentration

coming out of the digesters

(g) Assuming the belt press solids capture is 90%; if it produces a 20% cake which is 68 lbs/ft³

in how many hours will it take to ﬁll a 3 yd³bin with the cake?

Solution:

(a) What is the combined solids concentration?

Combined solids concentration: C₁∗V₁+C₂∗V₂= C₃∗V₃

C₁∗V₁+C₂∗V₂C₁∗V₁+C₂∗V₂

4.5∗60,000+5∗28,000

= 4.66%

=⇒ C₃=

V₁+V₂

60,000+28,000

60,000

lbs TS from primary sludge: (4.5∗10,000) TS∗

MGD∗8.34 = 22,518 lbs TS

MGD∗8.34 = 11,676 lbs TS

1,000,000

28,000

1,000,000

lbs TS from secondary sludge: (5∗10,000) TS∗

Total TS = 22,518+11,676=34,194 lbs TS

34,194 lbs TS

TS conc.(mg/l) =

= 46,591

or 4.66%

88,000

1,000,000

8.34∗

MGD

(b) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of

VS are in the combined sludge?

lbs of VS in combined sludge: C_VS1∗V₁+C_VS2∗V₂= C_VS3∗V₃

C_VS1∗V₁+C_VS2∗V₂C_VS1∗V₁+C_VS1∗V₂

=⇒ C_VS3

V₁+V₂

4.5∗60,000∗0.68+5∗28,000∗0.83

= 3.406%

60,000+28,000

88,000

lbs VS = 3.41∗10,000mg/l ∗

∗8.34 = 25,027 lbs VS

1,000,000

60,000

1,000,000

lbs VS from primary sludge: (4.5∗0.68∗10,000) VS∗

MGD∗8.34 = 15,312 lbs VS

28,000

lbs VS from secondary sludge: (5 ∗ 0.83 ∗ 10,000) VS ∗

MGD ∗ 8.34 =

1,000,000

9,691 lbs VS Total VS = 15,312+9,691= 25,003 lbs VS

lbs VS/ft³-day?

22.27 Dewatering math problems

445

lbs VS per −day

Digester volume ( ft³)

(d) If the digester VS reduction is 60%, what is the gas production (ft³/day) if the digester

produces 14 ft³gas/lb VS destroyed?

25,003 lbs VS per −day

0.785∗120²∗20 ft³

0.11 lbs VS

day− ft³

VS loading =

Digester gas production = lbs VS reduced∗gas production rate(ft³/lb lbs VS reduced

14 ft³gas

lb VS destroyed

=⇒ 25,003∗0.6 lbs VSreduced ∗

= 210,025 ft³gas per −day

(e) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

are operated at a maximum ﬂow of 50 GPM?

Flow to the belt press=Flow to the digesters as digesters are ﬁxed volume tanks

=⇒ ﬂow in to the digester = ﬂow out of the digesters - which will be the ﬂow to the belt press

gal

day

day 1440 min

=⇒ @50 GPM belt press flow capacity need 2 belt presses

gal

min

=⇒ ﬂow to the belt press = 88,000 gal =⇒ 88,000

∗

= 61

446

Chapter 22. Water Math

(f) Knowing the digester VS reduction and the TS feed, calculate the solids (TS) concentration

coming out of the digesters

Approach: Calculate the lbs of VS destroyed in the digester and subtract that out of the lbs

TS feed to the digester. This will be the mass of solids leaving the digester. The mass of

solids divided by the volume of the feed (also the out) would be the TS concentration

34,194 lbs TS feed −25,003∗0.6 lbs VS reduced

=⇒

lbs

10⁶lbs

88,000 gal ∗8.34

∗

gal 1,000,000lbs

lbs solids

= 20,441

or 26,150 ppm or 2.61%

10⁶lbs

(g) Assuming the belt press solids capture is 90%; if it produces a 20% cake which is 68 lbs/ft³

in how many hours will it take to ﬁll a 3 yd³bin with the cake?

Approach

i. Calculate the lbs of solids feed to the belt press.

ii. Based on the 90% capture rate, calculate the lbs of solids leaving the belt press.

iii. Knowing the cake is 20% solids, calculate the lbs of cake produced.

iv. Knowing the density of the cake (68 lbs/ ft³), calculate the volume of the cake produced.

v. Knowing the cake volume produced, calculate the time it will take to ﬁll the 3 cu. yd bin

Calculations:

i. (34,194 lbs TS feed −25,003∗0.6 lbs VS reduced) solids feed belt press

= 19,192 lbs TS solids feed

ii. solids leaving belt press: 19,192*0.9=17,273 lbs

lbs solids

day

lb cake

lbs cake

day

iii. lbs cake : 17,273

∗

= 86,365

0.2 lb solids

lbs cake

ft³

yd³

day

iv. Volume of cake produced: 86,365

∗

day

68 lbs cake 27 ft³24 hrs

yd³cake

1.96

v. Time to ﬁll the 3 cu. yd bin: 3 yd³∗1.96

= 1.53 hrs

yd³cake

Question 1: A sludge thickened from 1% to 4% solids will be reduced in volume by how much?

Solution: Assume 1 lbs of solids are in the sludge.

100 lbs sludge

Sludge weight for 1% solids = 100 lbs because, 1% =⇒

1 lb solids

100 lbs sludge

4 lb solids

25 lbs sludge

1 lb solids

Sludge weight for 4% solids = 25 lbs because, 4% =⇒

Thus, sludge weight and volume of the 4% sludge will be 25% of the original 1% sludge

Alternatively, C_1(1%)∗V_1(1%)= C_2(4%)∗V_2(4%)=⇒ 0.01V_1(1%)= 0.04V_2(4%)=⇒ V_2(4%)

0.25V_1(1%)

Thus, volume of the 4% sludge will be 25% of the original 1% sludge.

39. If a belt ﬁlter press is used 14 hrs per day to dewater 90 GPM digester sludge feed with 2.7% solids,

the daily lbs solids feed is:

Solution:

lbs solids feed to belt press:

gal

min

lbs

gal

lbs solids

lb feed sludge

min

day

lbs solids

day

∗8.34

∗0.027

∗(14∗60)

= 17,024

22.27 Dewatering math problems

447

40. A belt ﬁlter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.

The belt press produces 22% biosolids (bulk density - 70 lbs/ft³) which is loaded into trucks for

offsite disposal. Five truckloads of biosolids are produced every three days with each truckload

averaging 13.3 yd³of material.

(a) Calculate the lbs of solids feed to the belt press (5 points)

(b) Calculate lbs cake hauled by the trucks over the three days (5 points)

(d) Calculate percent solids recovery (lbs solids produced per lbs feed solids) (5 points)

Solution:

(a) lbs solids feed to belt press:

gal

min

lbs

gal

lbs solids

lb feed sludge

min

day

lbs solids

day

∗8.34

∗0.025

∗(10∗60)

= 10,008

(b) lbs cake hauled by truck over three days:

yd³sludge

truck

trucks

three days

ft³

yd³

lbs cake

ft³sludge

lbs cake

three days

13.3

∗5

∗27

∗70

= 125,685

lbs cake

three days

lbs solids

lbs cake

lbs solids

three days

125,685

∗0.22

= 27,651

(d) Percent solids recovery:

lbs solids

27,651

lbs solids in cake

three days

lbs solids

= 0.92 or 92%

lbs solids fed to belt press

10,008∗3

three days

41. Lab data from your 100,000 gallon primary anaerobic digester, which receives primary sludge only,

is shown below. Using this data :

(a) Calculate the average volatile solids reduction. Compare your calculated value to generally

accepted ranges for a healthy anaerobic digester. Comment.

(b) Compare the other data to expected ranges.

steps that may be taken to mitigate the problem.

(d) Should slake lime be added ? Why or why not ?

Data

CO₂

Date

Alkalinity

(mg/l)

Volatile

acids

Raw

Digested

Sludge

(mg/l)

280

(TS%)

5.4

(VS%)

65.5

(VS%)

9/02

9/09

9/16

9/17

7.10

7.00

6.90

6.85

3,200

3,020

2,800

2,720

35.5

36.0

37.7

38.2

320

5.0

66.7

53.8

54.2

400

4.9

65.9

450

42. Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and

23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?

Solution:

Correct Answer:

C₁∗V₁+C₂∗V₂= C₃∗V₃

7.48gal

4.9∗8,000 ft³∗

+5.2∗23,000

ft³

C₁∗V₁+C₂∗V₂

V₁+V₂

=⇒ C₃=

8000∗7.48+23,000

4.98% −Correct Answer

448

Chapter 22. Water Math

Incorrect Answer #1

C₁∗V₁+C₂∗V₂= C₃∗V₃

C₁∗V₁+C₂∗V₂

V₁+V₂

4.9∗8,000ft³+5.2∗23,000

=⇒ C₃=

= 5.12% −

8000+23,000

Incorrect Answer#1

Incorrect Answer #2

6.23%

Incorrect Answer #3

3.84%

43. You have two 110 foot diameter primary clariﬁers. The raw wastewater ﬂow of 1.5 MGD is divided

equally between the these two basins. Raw settled primary sludge is pumped continuously from

both clariﬁers for thickening in a single 40 foot diameter gravity thickener. The average suspended

solids are 210 mg/L for the raw inﬂuent ﬂow and 85 mg/L for the primary efﬂuent. Calculate the

average solids loading on the thickener in pounds/ft²/day Ans. 1.25lbs/ft²/day

44. A belt ﬁlter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.

45. A belt ﬁlter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.

The belt press produces 22% biosolids (bulk density - 70lbs/cu. ft) which is loaded into trucks for

offsite disposal. Five truckloads of biosolids are produced every three days with each truckload

averaging 13.3 cu. yd of material.

(a) Calculate the lbs of solids feed to the belt press (5 points)

(b) Calculate lbs cake hauled by the trucks over the three days (5 points)

(d) Calculate percent solids recovery (lbs solids produced per lbs feed solids) (5 points)

(e) Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and

23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?

a. 3.84

b. 6.23

c. 5.12

*d. 4.98

46. 325 GPM of WAS with 7,000 mg/l TSS is thickened in a DAFT. Assuming an air:solids ratio of

0.05 lb of air for each lb of feed solids, The SCFM (standard cubic feed per minute) air required to

thicken the WAS given 1 SCFM = 0.075 lb is:

*a. 12.6 SCFM

b. 0.126 SCFM

c. 12,600 SCFM

d. 1.26 SCFM

47. When you spread sludge on agricultural land, the annual application rate of cadmium in the sludge

should be less than 2 lbs/acre/year If your sludge contains 30 mg cadmium per kilogram of solids

and your plant produces 950,000 lbs per year of dry solids, how many acres do you need?

a. 3 acres

*b. 14 acres

c. 19 acres

d. 27 acres

48. 42,000 gallons of 6% sludge containing 67% volatile matter is pumped to the digester. The di-

22.27 Dewatering math problems

449

gester reduces the volatile matter by 52%. What volume of sludge in gallons containing 5% solids

remains after digestion? [Ans: 32,841 gal]

49. An Imhoff cone result is 5.5 ml/l. Approximately how many gallons of primary sludge will need to

be pumped if the ﬂow is 0.7 MGD? [3,850 gallons]

50. A sludge digester equipped with a ﬂoating cover is 31 ft. inside diameter. The corbels are 16 feet

above the ﬂoor and the top of the wall is 20 feet above the ﬂoor. The digester currently has a sludge

depth of 16.5 ft. How many gallons of sludge are needed to displace the cover 1 foot? [5,643

galllons]

51. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of

the VS are destroyed, creating 10 ft.3 of gas per lbs. of VS destroyed. How much gas is produced

each day? [Ans. 11,676 cu. ft/day]

52. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.

secondary sludge at 1.3% solids. What is the combined solids concentration? [Ans. 2.28%] If the

primary sludge is 68% VS and the secondary sludge is 63% VS, how many pounds of VS are in the

combined sludge? [Ans. 5400.3 lbs VS]

53. 12,000 gallons of 1.8% sludge is pumped to a thickener, and is thickened to 3800 gallons of 4.9%

solids. The supernatant is returned back to the head of the STP for treatment. What is the volume in

gallons and solids concentration in ppm of the supernatant?[Ans. 24,980 gallons]

54. How many pounds of solids are pumped to a digester each day if the digester receives 10,000 gpd

of sludge at 5% solids concentration?

Solution:

lbs TS 10,000gal sludge (8.34∗0.05lbsTS)

= 4,170

lbs TS

day

∗

day

gal sludge

55. Calculate the % VS reduction in a digester given the volatile solids content of the inﬂuent sludge to

the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%

0.7−0.525

Solution: Digester VS reduction(%) =

∗100 = 53%

0.7−0.7∗0.525

56. Calculate the VS loading to the digester in lbs/day if 10,000 gallons of sludge containing 5% TS

with and average VS content of 78%

Solution:

Digester VS loading (lbs/day)

10,000 gallons sludge 8.34lbs sludge 0.05∗0.78lbsVS

∗

= 3,253lbs sludge per day

day

gal

lb sludge

57. The sludge feed to a digester is 12,000 ft³/day. The sludge contains 4.5% total solids with 70%

volatile solids. What is the digester loading in lbs VSS/day per ft³if the digester diameter is 100 ft

and a working sludge level of 20 ft.

Solution:

lbs VS

Digester loading

day

Digester volatile solids loading Rate =

Digester volume(V)ft³

450

Chapter 22. Water Math

ft³sludge

day

gal

ft³

lbsVS

gal sludge

12,000

∗7.48

∗(8.34∗0.045∗0.70)

lbs VS

= 0.15

day− ft³

(

∗100²∗20)ft³

58. How much gas is produced in the above digester in ft³/day if the digested sludge contains 2.5%

total solids of which 59% is volatile solids and the gas production rate is 14 ft³/lb VS destroyed?

Solution:

0.70−.59

Digester VS reduction(%) =

∗100 = 38.3%

0.70−0.70∗0.59

lbs VS reduction 3153lbs VS feed ∗0.383 VS reduction

lbs VS reduction

day

= 1,208

day

ft³gas produced

day

lbs VS reduced 14 ft³gas produced

ft³digester gas produced

= 1208

∗

= 16,912

day

lb VS reduced

day

59. The sludge feed to a digester is 80,000 gal/day. The sludge contains 4.5% total solids with 75%

volatile solids. If 50% of VS are reduced in the digester:

(a) Find lbs VS destroyed per 1000 gal of digester capacity per day if the digester radius is 55 ft

with an operating sludge level of 25 ft.(5 points)

Solution:

gal

ft³

Digester Volume: (π ∗55²∗25) ft³∗7.48

= 1,776,220gallons = 1,776.2 (1000 gallons)

lbs

gal

lbsVS

lbs VS reduction

∗0.5

lb VS

80,000gal ∗8.34

∗(0.045∗0.75)

lbs VS reduction

day

lbs VS reduction

day

= 11,259

lbs VS reduction

day

1,776.2 (1000 gallons)

11,259

lbs VS reduction

1000 digester capacity

lbs VS reduction

1000gallons digester volume

= 6.3

(b) What is the digester gas production in Btu/day? Assume 14 cu. ft digester gas per lb of VS

destroyed and a 650 Btu/cu. ft heating value for the digester gas produced (5 points)

Solution:

ft³gas produced

day

lbs VS reduced 14 ft³gas produced

ft³digester gas produced

day

= 11,259

∗

= 157,626

day

lb VS reduced

BTU produced

ft³gas produced 650BTU gas produced

BTU produced

= 157,626

∗

= 102,456,900

day

ft³gas

day

60. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.

secondary sludge at 1.3% solids. What is the combined solids concentration? If the primary sludge

is 68% VS and the secondary sludge is 63% VS, what is the VS concentration (%) in the combined

sludge?

22.27 Dewatering math problems

451

Solution:

Combined solids concentration: C₁∗V₁+C₂∗V₂= C₃∗V₃

C₁∗V₁+C₂∗V₂C₁∗V₁+C₂∗V₂

4.1∗15,000+1.3∗28,000

= 2.28%

=⇒ C₃=

V₁+V₂

15,000+28,000

Lbs of VS in combined sludge: C_VS1∗V₁+C_VS2∗V₂= C_VS3∗V₃

C_VS1∗V₁+C_VS2∗V₂C_VS1∗V₁+C_VS1∗V₂

=⇒ C_VS3

V₁+V₂

4.1∗15,000∗0.68+1.3∗28,000∗0.63

15,000+28,000

= 1.50%

452

Chapter 22. Water Math

61. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of

the VS are destroyed, creating 15 ft³of gas per lbs. of VS destroyed. How much gas is produced

each day?

lbs

gal

lbsVS 0.5lbs VS reduction

10,000gal ∗8.34

∗(0.04∗0.7)

∗

lbs VS reduction

day

lbs VS reduction

day

lb VS

= 1,168

day

ft³gas produced

day

lbs VS reduced 15 ft³gas produced

ft³digester gas produced

day

= 1,168

∗

= 17,520

day

lb VS reduced

62. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and

28,000 gal secondary sludge at 5% solids.

(a) What is the combined solids concentration? (3 points)

60,000∗4.5+28,000∗5

= 4.66%

88,000

(b) How many lbs of solids (per day) are in the combined sludge (3 points)

88,000gal ∗8.34∗0.0466 = 34,200lbs

VS are in the combined sludge? (3 points)

8.34∗0.045∗0.68lbs VS

8.34∗0.05∗0.83lbs VS

60,000gal∗

+28,000gal∗

= 25,003lbs VS

gal

(d) What is the digester feed VS%? (3 points)

25,003lbs VS

34,200lbsTS

= 73.1%

(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs

VS/ft3-day (3 points)

25,003lbs VS

0.785∗120²∗20

0.11lbs VS

day− ft³

(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)

22.27 Dewatering math problems

453

0.731−.52

Digester VS reduction(%) =

∗100 = 60.1%

0.731−0.731∗0.52

(g) What is the gas production (ft³/day) if the digester produces 14 ft³gas/lb VS destroyed (3

points)

lbs VSdestroyed 14 ft³gas produced

25,003*0.601

∗

= 210,375ft³gas

lb VS destroyed

(h) How many belt presses are needed to keep up with the digested sludge ﬂow if the belt presses

can be operated at a maximum ﬂow of 50 GPM (3 points)

gal sludge

day

1440min

88,000

∗

= 61GPM

@50 GPM per press - 2 BFP required

(i) If the digested sludge feed to the belt ﬁlter presses is 2.6% and assuming the belt press feed

solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)

gal

day

lbs solids feed

day

lbs solids captured

lbs solids feed

88,000

∗8.34∗0.026

∗0.9

= 17,174lbs solids per day

(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)

17,174lbs solids produced

100lbs cake

20lbs solids produced 2,000lbs

tons

42.9tons

day

∗

day

(k) If the cake density is 68 lbs/cu. ft, how much time will it take to ﬁll a 30 cu. yd bin (3 points)

(42.9∗2000)lbs cake

day

59.583

(Ans. 15.4 hours)

∗

min

day

1440min lbs cake

59.583

min∗

ft³

0.876 ft³

lbs cake

68lbs cake

min

27 ft³

yd³

min

hrs

30yd³∗

∗

0.876ft³60min

= 15.4hrs

454

Chapter 22. Water Math

(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)

tons $65

day ton

$2,789

day

42.9

∗

63. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS

with and average VS content of 76% is fed to the digester

Solution:

25,000 Gal 8.34 lbs sludge (0.045∗0.76) lbs VS feed

7,131 lbs VS

∗

day

Gal

lb sludge

day

64. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge

feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge

feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-

ft3 )

65. 10,000 gallons/day of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). If

50% of the VS is destroyed, how many lbs of VS is destroyed per day?

Solution:

10,000 Gal 8.34 lbs sludge 0.04∗0.7 lbs VS feed 0.5 lbs VS destroyed

1,168lbs VS destroyed

∗

day

Gal

lb sludge

lbs VS feed

day

66. Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and

23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?

a. 3.84

b. 6.23

c. 5.12

*d. 4.98

67. Calculate the VS loading to the digester in lbs/day if 10,000 cu. ft of sludge containing 4.8% TS

with and an average VS content of 82% (Ans: 24,554)

68. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS

with and average VS content of 76% is fed to the digester (Ans: 7,131)

69. Calculate the % VS reduction in a digester given the volatile solids content of the feed sludge to the

digester is 76% and the volatile solids content of the sludge leaving the digester is 55% (Ans: 61.4)

70. If an anaerobic digester receives sludge with VS of 76% and discharges digested sludge with a 56%

VS. Its VS reduction is:

a. 36%

b. 48%

*c. 60%

d. 89%

71. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS

with and average VS content of 76% is fed to the digester

Correct Answer(s): a. 7131.0

72. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge

feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge

feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-

ft3 )

Correct Answer(s):

73. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS

with and average VS content of 76% is fed to the digester

*a. 7,131 lbs/day

22.27 Dewatering math problems

455

b. 17,432 lbs/day

c. 19,256 lbs/day

d. 26,244 lbs/day

74. What is the digester efﬁciency if the volatile solids concentration entering an anaerobic digester is

70% and the volatile solids concentration leaving the digester is 50%?

*a. 29%

b. 40%

c. 57%

d. 71%

75. You are operating a 1 MGD trickling ﬁlter plant with a single stage anaerobic digester You pump

about 3,000 gpd of sludge to the ﬁxed cover digester. The supernatant has a BOD of 950 mg/L The

BOD loading coming from the supernatant in pounds per day is

a. 12

*b. 24

c. 36

d. 48

76. How many pounds of solids are pumped to a digester each day, if the digester receives 10,000 gpd

of sludge at 5% density?

a. 417 lbs/day

b. 1,497 lbs/day

c. 2,337 lbs/day

d. 3,273 lbs/day

*e. 4,170 lbs/day

77. If a digester loading is 0.05 lbs VSS/ft3/Day, how large must the digester be if it receives 1,000 lbs.

of suspended solids per day with a volatile percentage of 75%?

a. 3,750 cu.ft

b. 10,000 cu.ft

*c. 15,000 cu.ft

d. 26,666 cu.ft

e. 37,500 cu.ft

78. What is the digester efﬁciency if the volatile solids concentration entering an anaerobic digester is

70% and the volatile solids concentration leaving the digester is 50%?

*a. 29%

b. 40%

c. 57%

d. 71%

79. You are operating a 1 MGD trickling ﬁlter plant with a single stage anaerobic digester You pump

about 3,000 gpd of sludge to the ﬁxed cover digester. The supernatant has a BOD of 950 mg/L The

BOD loading coming from the supernatant in pounds per day is

a. 12

*b. 24

c. 36

d. 48

80. How many pounds of solids are pumped to a digester each day, if the digester receives 10,000 gpd

of sludge at 5% density?

456

Chapter 22. Water Math

a. 417 lbs/day

b. 1,497 lbs/day

c. 2,337 lbs/day

d. 3,273 lbs/day

*e. 4,170 lbs/day

81. If a digester loading is 0.05 lbs VSS/ft3/Day, how large must the digester be if it receives 1,000 lbs.

of suspended solids per day with a volatile percentage of 75%?

a. 3,750 cu.ft

b. 10,000 cu.ft

*c. 15,000 cu.ft

d. 26,666 cu.ft

e. 37,500 cu.ft

Disinfection Math Problems

1. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/l at a

5.0 MGD ﬂow.

Solution:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day = 12∗5∗8.34 = 500.4lbs/day

2. What is the chlorine demand if the chlorine dosage is 15 mg/L and the residual is 3 mg/l? Solution:

Chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine demand = chlorine dosage−chlorine residual = 15−3 = 12mg/l

3. If 80 pounds of chlorine are applied each day to a ﬂow of 1.5 MGD, what is the dosage in mg/l?

Solution:

Applying the pounds formula:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

=⇒ conc.(mg/l) =

= 6.4mg/l

1.5∗8.34

flow(MGD)∗8.34

4. How many pounds per day of chlorine will be required to disinfect a secondary efﬂuent ﬂow of

1.68 MGD if the chlorine demand is found to be 8.5 mg/l and a residual of 3 mg/l is desired? Chlo-

rine dosage = chlorine demand + chlorine residual

chlorine dosage = 8.5+3 = 11.5mg/l

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34 = 1.68∗11.5∗8.34 = 161.2lbs/day

5. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a ﬂow of 2.8

MGD, how many pounds per day should the chlorinator be set to deliver.

Solution:

Chlorine dosage = chlorine demand + chlorine residual

chlorine dosage = 4.8+0.75 = 5.55mg/l

To calculate pounds per day, applying the pounds formula:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34 = 2.8∗5.55∗8.34 = 129.6lbs/day

6. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

22.27 Dewatering math problems

457

d. 8.3 mg/L

Solution:

Chlorine dosage(lbs/day) = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

=⇒ chlorine dosage conc.(mg/l) =

= 7.5mg/l

1.2∗8.34

flow(MGD)∗8.34

Chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine demand = chlorine dosage−chlorine residual = 7.5−2.6 = 4.9mg/l

7. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a

wastewater efﬂuent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed

at this dosage to a ﬂow of 0.5 MGD?

Solution:

Chlorine dosage(lbs/day) = conc.(mg/l)∗ flow(MGD)∗8.34 = 24∗0.5∗8.34 = 100lbs/day

8. 25 lbs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate the

chlorine dosage in mg/l.

Solution:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

=⇒ conc.(mg/l) =

= 12mg/l

0.25∗8.34

flow(MGD)∗8.34

9. You wish to dose the inﬂuent channel at 5 mg/l chlorine to help control odors. The ﬂow is 11.5

MGD.. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine

residual to control odors?

Solution:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34 = 5∗11.5∗8.34 = 480lbs/day

No. Chlorine residual is for disinfection only. Odor control would not require a residual

10. Jar testing shows that the chlorine demand of an efﬂuent is 12.5 mg/l. In order to assure disinfec-

tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to

assure disinfection?

Solution: chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine dosage = (12.5 + 1)mg/l = 13.5mg/l

lbs/day = 13.5(mg/l)∗1(MGD)∗8.34 = 112.6lbs/day

11. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily ﬂow is

3.5 MGD? What is the chlorine demand if the residual is 1.3 mg/l?

Solution:

a. Chlorine dosage(lbs/day) = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

flow(MGD)∗8.34

120

3.5∗8.34

=⇒ chlorine dosage conc.(mg/l) =

= 4.1mg/l

b. Chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine demand = chlorine dosage−chlorine residual = 4.1−1.3 = 2.8mg/l

12. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a

wastewater efﬂuent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed

at this dosage to a ﬂow of 0.5 MGD?

Solution:

Chlorine dosage(lbs/day) = conc.(mg/l)∗ flow(MGD)∗8.34 = 24∗0.5∗8.34 = 100lbs/day

13. 25 lbs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate the

chlorine dosage in mg/l.

Solution:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

=⇒ conc.(mg/l) =

= 12mg/l

0.25∗8.34

flow(MGD)∗8.34

458

Chapter 22. Water Math

14. You wish to dose the inﬂuent channel at 5 mg/l chlorine to help control odors. The ﬂow is 11.5

MGD.. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine

residual to control odors?

Solution:

lbs/day = conc.(mg/l)∗ flow(MGD)∗8.34 = 5∗11.5∗8.34 = 480lbs/day

No. Chlorine residual is for disinfection only. Odor control would not require a residual

15. Jar testing shows that the chlorine demand of an efﬂuent is 12.5 mg/l. In order to assure disinfec-

tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to

assure disinfection?

Solution: chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine dosage = (12.5 + 1)mg/l = 13.5mg/l

lbs/day = 13.5(mg/l)∗1(MGD)∗8.34 = 112.6lbs/day

16. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily ﬂow is

3.5 MGD? What is the chlorine demand if the residual is 1.3 mg/l?

Solution:

a. Chlorine dosage(lbs/day) = conc.(mg/l)∗ flow(MGD)∗8.34

lbs/day

flow(MGD)∗8.34

120

3.5∗8.34

=⇒ chlorine dosage conc.(mg/l) =

= 4.1mg/l

b. Chlorine dosage = chlorine demand + chlorine residual

=⇒ chlorine demand = chlorine dosage−chlorine residual = 4.1−1.3 = 2.8mg/l

17. A 2.5 MGD secondary ﬂow is disinfected by the application of 320 lbs of chlorine per day. This

dose provides a chemical residual of 2.1 mg/l. There is a need to switch to the use of sodium

hypochlorite which has a 12.5% available chlorine, SG of 1.2 and a cost of $0.60 per gallon. Chlo-

rine costs $0.28/lb.

Calculate: 1) The chlorine demand, and 2) Cost difference ($ per day) between chlorine and

sodium hypochlorite

Dosage = Demand + Residual

Dosage:

320lbs chlorine

= 2.5MGD∗8.34∗x

day

x^mg

= 15.34^mg

320

2.5∗8.34

Chlorine Demand = Dosage - Residual = 15.34 - 2.1 = 13.24

$320 $0.28

Cost per day to use chlorine:

∗

= $89.60

To calculate the hypochlorite we need to determine the gallons per day of bleach required.

bleach

day

320^lbs

= x^gal

∗ 8.34∗1.2 _per

_bleach∗ 0.125^lbs

chlorine

day

lbs bleach

chlorine

gal

lb bleach

bleach

day

bleach

→ x^gal

= 256^gal

320

8.34∗1.2∗0.125

$0.60

day

256^gal

∗

= $153.48 Cost difference $153.48 - $89.60 = $63.88

gal bleach

bleach

day

22.27 Dewatering math problems

459

18. Three hundred pounds (300 lbs) per day of chlorine (cost = $0.32/lb) are being used to disinfect a

secondary efﬂuent ﬂow of 6.5 MGD. Due to safety concerns, the substitution of sodium hypochlo-

rite (12.5% available chlorine, 10.0 lbs/gal and a cost of $0.55/gal) for the chlorine is being consid-

ered at this plant. What is the cost difference per day between the chlorine and sodium hypochlo-

rite? (Ans:$36/year)

19. The operator at a 1.5 MGD conventional activated sludge plant is considering using either HTH or

sodium hypochlorite as an alternative to chlorine gas. Currently chlorine is being dosed at 15 mg/L

in order to achieve a residual of 3.0 mg/L. Using the data provided below calculate the daily cost

for chlorine, HTH, and sodium hypochlorite (NaOCl) (Sp.Gravity 1.21).

Chlorine → 0.15 $/lb

HTH (70% available chlorine) → 0.25 $/lb

NaOCl (15% available chlorine) → 0.35 $/gal

SOLUTION

lbs chlorine required:

15mg chlorine

1.5MG 8.34lbs

188lbs chlorine

∗

day

gallon

day

Daily cost if chlorine is used:

$0.15

chlorine

188lbs chlorine∗ _lb

= $28.20

Daily cost if HTH is used:

lb HTH

$0.25

188lbs chlorine∗ _0.7lb

∗

= $67.14

chlorine lb chlorine

Daily cost if NaOCl is used:

188lbs chlorine∗ _0.15lb

gal NaOCl

chlorine 8.34∗1.21lbs NaOCl gal NaOCl

lb NaOCl

$0.35

∗

= $43.47

460

Chapter 22. Water Math

20. A water storage tank is 30 feet in diameter and has a water depth of 18.5 feet. It is desired to super-

chlorinate this tank with 30 ppm of chlorine, how many pounds of HTH will be required (HTH has

70% available chlorine)?

Tank Volume:

0.785∗30²∗18.5 ft³∗7.48^g_f^a_t^l= 97,765gal

lbs HTH required:

30lbs chlorine

8.34lbs water

lb HTH

∗

∗97,765gal water ∗ _0.70lb

= 35lbs HTH

chlroine

1,000,000lbs water

gal water

21. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a ﬂow of 2.8

MGD, how many pounds per day should the chlorinator be set to deliver.

*a. 130

b. 116

c. 112

d. 18

e. 16

22. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

23. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5

mg/L, approximately how many pounds of chlorine per day will be required to treat a ﬂow of 5.0

MGD?

a. 21 lbs

b. 792 lbs

c. 813 lbs

*d. 834 lbs

24. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

25. If 25 Ibs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate

the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.

a. 2.5 mg/L

*b. 10.8 mg/L

c. 15.1 mg/L

d. 6.3 mg/L

26. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at

6.0 MGD ﬂow

a. 60 lbs/day

b. 450 lbs/day

22.27 Dewatering math problems

461

c. 500 lbs/day

*d. 600 lbs/day

e. 6,700 lbs/day

27. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

28. Chlorine is being fed at the rate of 200 pounds per day. Plant ﬂow is 4 MGD. The chlorine residual

is measured and found to be 3 mg/L Calculate chlorine demand.

a. 3 mg/L

*b. 2 mg/L

c. 16 mg/L

d. 9 mg/L

29. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a ﬂow of 2.8

MGD, how many pounds per day should the chlorinator be set to deliver.

*a. 130

b. 116

c. 112

d. 18

e. 16

30. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

31. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5

mg/L, approximately how many pounds of chlorine per day will be required to treat a ﬂow of 5.0

MGD?

a. 21 lbs

b. 792 lbs

c. 813 lbs

*d. 834 lbs

32. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

33. If 25 Ibs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate

the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.

a. 2.5 mg/L

*b. 10.8 mg/L

462

Chapter 22. Water Math

c. 15.1 mg/L

d. 6.3 mg/L

34. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at

6.0 MGD ﬂow

a. 60 lbs/day

b. 450 lbs/day

c. 500 lbs/day

*d. 600 lbs/day

e. 6,700 lbs/day

35. What is the chlorine residual of a 2.5 MGD secondary efﬂuent stream if its chlorine demand is 9

mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4

36. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD ﬂow?

Ans: 500.4

37. What is the chlorine residual of a 2.5 MGD secondary efﬂuent stream if its chlorine demand is 9

mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4

38. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD ﬂow?

Ans: 500.4

39. What is the chlorine residual of a 2.5 MGD secondary efﬂuent Ans: 5.4

40. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD ﬂow?

Ans: 500.4

41. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a

wastewater efﬂuent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed

at this dosage to a ﬂow of 0.5 MGD? Ans: 100.0

42. 25 lbs/day of chlorine is being applied to a wastewater efﬂuent ow of 250,000 gpd. Calculate the

chlorine dosage in mg/l. Ans: 12.0

43. You wish to dose the in inﬂuent channel at 5 mg/l chlorine to help control odors. The ﬂow is 11.5

MGD. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine

residual to control odors? Ans:480.0

44. Jar testing shows that the chlorine demand of an efﬂuent is 12.5 mg/l. In order to assure disinfec-

tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to

assure disinfection? Ans: 112.6

45. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily ﬂow is

3.5 MGD? Ans: 4.1

46. What is the chlorine residual of a 2.5 MGD secondary efﬂuent stream if its chlorine demand is 9

mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4

47. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD ﬂow?

Ans: 500.4

48. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

49. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

22.27 Dewatering math problems

463

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

50. Chlorine is being fed at the rate of 200 pounds per day. Plant ﬂow is 4 MGD. The chlorine residual

is measured and found to be 3 mg/L Calculate chlorine demand.

a. 3 mg/L

*b. 2 mg/L

c. 16 mg/L

d. 9 mg/L

51. Chlorine is being fed at the rate of 200 pounds per day. Plant ﬂow is 4 MGD. The chlorine residual

is measured and found to be 3 mg/L Calculate chlorine demand.

a. 3 mg/L

*b. 2 mg/L

c. 16 mg/L

d. 9 mg/L

52. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a ﬂow of 2.8

MGD, how many pounds per day should the chlorinator be set to deliver.

*a. 130

b. 116

c. 112

d. 18

e. 16

53. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

54. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5

mg/L, approximately how many pounds of chlorine per day will be required to treat a ﬂow of 5.0

MGD?

a. 21 lbs

b. 792 lbs

c. 813 lbs

*d. 834 lbs

55. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

56. If 25 Ibs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate

the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.

a. 2.5 mg/L

464

Chapter 22. Water Math

*b. 10.8 mg/L

c. 15.1 mg/L

d. 6.3 mg/L

57. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at

6.0 MGD ﬂow

a. 60 lbs/day

b. 450 lbs/day

c. 500 lbs/day

*d. 600 lbs/day

e. 6,700 lbs/day

58. How many pounds of chlorine will be used in one day if the ﬂow is 700,000 gpd and a uniform

dose of 12 mg/L is applied?

*a. 7 lbs

b. 15 lbs

c. 22 lbs

d. 26 lbs

59. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5

mg/L, approximately how many pounds of chlorine per day will be required to treat a ﬂow of 50

MGD?

a. 21 lbs

b. 792 lbs

c. 813 lbs

*d. 834 lbs

60. How many pounds of chlorine gas is necessary to treat 4,000,000 gallons of wastewater at a dosage

of 2 mg/L?

a. 61 lbs

b. 65 lbs

*c. 67 lbs

d. 69 lbs

61. If you need a chlorine residual of 1 mg/l , how many pounds of chlorine must be applied each day

if the ﬂ ow is 2.5 MGD and the chlorine demand is l5 mg/l?

a. 291 .pounds/day

b. 312 pounds/day

*c. 334 pounds/day

d. 419 pounds/day

e. 516 pounds/day

62. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

63. Chlorine is being fed at the rate of 200 pounds per day. Plant ﬂow is 4 MGD. The chlorine residual

is measured and found to be 3 mg/L Calculate chlorine demand.

a. 3 mg/L

22.27 Dewatering math problems

465

*b. 2 mg/L

c. 16 mg/L

d. 9 mg/L

64. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a ﬂow of 2.8

MGD, how many pounds per day should the chlorinator be set to deliver.

*a. 130

b. 116

c. 112

d. 18

e. 16

65. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

66. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5

mg/L, approximately how many pounds of chlorine per day will be required to treat a ﬂow of 5.0

MGD?

a. 21 lbs

b. 792 lbs

c. 813 lbs

*d. 834 lbs

67. Chlorine is being fed at the rate of 75 pounds per day. Plant ﬂow is 1.2 MGD. The chlorine residual

is measured and found to be 2.6 mg/L Calculate chlorine demand.

*a. 4.9 mg/L

b. 5.7 mg/L

c. 7.5 mg/L

d. 8.3 mg/L

68. If 25 Ibs/day of chlorine is being applied to a wastewater efﬂuent ﬂow of 250,000 gpd. Calculate

the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.

a. 2.5 mg/L

*b. 10.8 mg/L

c. 15.1 mg/L

d. 6.3 mg/L

69. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at

6.0 MGD ﬂow

a. 60 lbs/day

b. 450 lbs/day

c. 500 lbs/day

*d. 600 lbs/day

e. 6,700 lbs/day

466

Chapter 22. Water Math

Chemical Dosing Math Problems

1. Polymer is being added at 0.2 mg/l in order to achieve a 98% capture efﬁciency for a belt press.

The feed to the belt press is 75 gallons per minute, containing 2.5% solids. Given the polymer costs

$250 per gallon of 4.5% active polymer with a speciﬁc gravity of 1.08. What is the cost of polymer

per dry ton of solids captured

Solution:

lbs polymer required:

sludge

lbs sludge

0.2lbs polymer

75∗1440^gal

∗8.34

∗

day

gal sludge 1,000,000lbs sludge

polymer

day

= 0.1801^lbs

gallons polymer solution required:

polmyer

day

polymer solution

8.34∗1.08lbs polymer solution

lbs polymer

polymer solution

0.1801^lbs

= x^gal

∗

∗ 0.045_lb

day

gal polymer solution

polymer solution

day

=0.444^gal

Polymer cost:

0.444gal polymer soultion

$250

∗

gallon polymer soultion

day

$111

day

Dry tons of solids captured:

sludge

0.98 lbs solids captured

75∗1440^gal

∗

8.34∗0.025 lbs solids

ton solids

2000lbs solids

day

gal sludge

lbs solids

11tons dry solids

day

Polymer cost per dry ton of solids captured:

$111perday

= $10.09

11tons dry solids per day

2. A 50 MGD ﬂow is being treated with 20 mg/L ferric chloride. How many lbs of ferric chloride is

required daily

lbs ferric chloride required:

8,340lbs ferric chloride

20mg ferric chloride

50MG 8.34lbs

∗

day

gallon

day

3. If the ferric chloride solution used contains 40% dry ferric chloride with a speciﬁc gravity of 1.4,

what is its required feed rate in GPM (7 points)

Required FeCl₃feed (gal/min) to feed 8,340 lbs ferric chloride:

8,340lbs FeCl

xgal FeCl soltn. 8.34∗1.4lbs FeCl soltn

0.4lbs FeCl

1440min

day

∗

day

minute

gal FeCl soltn.

lbs FeCl soltn.

1.24gal

8,340

8.34∗1.4∗0.4∗1440

=⇒ x =

min

4. What is the daily ferric chloride dosing cost if the ferric chloride cost is $580/dry ton ferric chloride

(5 points)

Daily ferric chloride dosing cost:

$2,419

day

8,340lbs FeCl

ton

2000lbs ton FeCl

$580

∗

day

22.27 Dewatering math problems

467

5. A 0.5% (based on dry weight) solution of polymer is being fed to a secondary efﬂuent prior to sand

ﬁltration. It is desired to dose this efﬂuent at 2.5 mg/L. Assuming an efﬂuent ﬂow of 3000 gpm, at

what rate (gpm) should the polymer feed pump be set?

· lbs polymer required:

3000gallons 8.34 lbs e f fluent

2.5 lbs polymer

0.0626lbs polymer

∗

1,000,000lbs e f fluent

min

gallon

min

Required pumping rate to feed 0.0626 lbs polymer minute:

0.0626 lbs polymer

gallon polymer solution 8.34 lbs polymer solution

0.005 lbs polymer

lbs polymer solution

∗

min

minute

gallon polymer solution

1.5 gallon

gallon polymersolution

0.0626

8.34∗0.005

minute

min

6. How many pounds of dry polymer should be added to how many gallons of water to make enough

2% polymer solution to dose a 10 MGD secondary efﬂuent ﬂow at 3.0 ppm of polymer?

lbs polymer required:

8.34lbs e f fluent 3mg polymer

250.2lbs polymer

10MGD∗

∗

gallon

day

Required pumping rate to feed 250.2 lbs polymer minute:

250.2lbs polymer

xgallon polymer solution 8.34lbs polymer solution

0.02lbs polymer

lbs ploymer solution

∗

day

gallon polymer solution

1500gallon

xgallon polymer solution

day

250.2

8.34∗0.02

day

This can also be done as follows:

1500gallon

250.2lbs polymer lbs ploymer solution

gallon polymer solution

∗

8.34lbs polymer solution

day

0.02lbs polymer

day

So dissolve 250.2 lbs polymer in 1500 gallons

250.2lbs polymer

1500∗8.34lbs polymer solution

0.02lbs polymer

= 2%polymer

lb polymer solution

Check:

7. A 0.35% solution of polymer is being fed to a secondary efﬂuent prior to sand ﬁltration. The poly-

mer feed pump is set to pump 0.5 gpm to an efﬂuent ﬂow of 4200 gpm. What is the polymer dose

rate in ppm?

Pounds polymer pumped:

0.5gal PS

0.0146lbs Polymer

8.34lbs PS 0.0035lbs

∗

min

gal PS

lb PS

min

Polymer dose rate:

lbs polymer

_{e f fluent}(ppm or ^mg) =

= 0.42ppm polymer

0.0146lbs min

8.34∗4,200

10 lbs

10 lbs e f fluent

1,000,000

8. Liquid alum (49% alum, sp. gravity 1.32, $1.85/gal)) is being used to remove phosphorus from

a 600,000 gpd activated sludge efﬂuent. Two hundred milligrams per liter (200 mg/L) of alum,

Al₂(S0₄)₃.14H₂0, is required to give adequate removal of the phosphorus in this efﬂuent. Cal-

culate the daily cost of liquid alum needed to remove phosphorus. [Formula Weights: Al =27,

Al₂(S0₄)₃.14H₂0 =594]

468

Chapter 22. Water Math

lbs alum required:

200mg alum

0.6MG 8.34lbs

1001lbs alum

∗

day

gallon

day

Required liquid alum feed (gal/day) to feed 1001 lbs alum:

xgallon liquid alum 8.34∗1.32lbs liquid alum

1001lbs alum

0.49lbs alum

lbs liquid alum

∗

day

minute

gallon liquid alum

1001

8.34∗1.32∗0.49

=⇒ x =

= 186gal

Daily cost of liquid alum to remove phosphorous:

$344.10

day

186gal

day

$1.85

gal

∗

22.27 Dewatering math problems

469

9. A 1.5% polymer solution (based on dry weight) is to be fed at the rate of 3.5 pp to a secondary

efﬂuent ﬂow of 4.0 MGD. (a) Calculate the polymer pump feed rate (gallon per minute) necessary

to dose this secondary efﬂuent. (b) How many gallons per day of polymer solution will be required

for an average ﬂow of 3470 gpm?

(a)

• Polymer required:

116.8lbs polymer

4∗3.5∗8.34 =

day

gal

• Feed rate ( ):

min

gal

min

116.8lbs polymer 100lbs polymer solution

gal polymer solution

= 0.65

8.34lbs polymer solution 1440min

day

∗

day

1.5lbspolymer

(b)

• Polymer required:

ꢄ

ꢅ

3470gal

145.9lbs polymer

1,000,000gal

1440min

day

∗

MGD∗3.5∗8.34 =

min

day

• Feed rate (_m^ga_in^l):

gal

min

145.9lbs polymer 100lbs polymer solution

gal polymer solution

= 0.81

8.34lbs polymer solution 1440min

day

∗

day

1.5lbspolymer

10. Liquid alum (Sp. gravity 1.32, 49% Alum, $1.70 per gallon,) is being used to remove phosphorus

from a 595,000 gal/day secondary. A dose of fourteen milligrams per liter (14.0 mg/L) of alu-

minum is required to give adequate removal of the phosphorus in this efﬂuent. Calculate the daily

cost of liquid alum. (Note: Dry alum contains 9.4% aluminum).

Aluminum Required:

69.47lbs aluminum

0.595∗14∗8.34 =

day

Cost of alum (_day):

gal alum solution

(8.34∗1.32)lbs alum solution

69.47lbs aluminum

100lbs alum

100lbs alum solution

49lbs alum

$1.70

∗

gal alum solution

day

9.4lbs aluminum

$232.91

day

11. Polymer solution (0.5% weight to weight) is being fed at the rate of 0.83 gpm to a secondary efﬂu-

ent ﬂow of 1950 gpm prior to sand ﬁltration. Calculate the polymer dose in units of ppm.

lbs polymer

10 lbs e f fluent

Polymer Dose -

which is

or ppm:

polymer solution(PS) & polymer(P)

2.13mg

0.83gal PS

8.34lbs PS 0.005lbs

min

∗

polymer

(8.34∗1950)

1,000,000

min

gal PS

lbs PS

10 lbs e f fluent

12. A 4.5% (weight to weight basis) solution of polymer is being fed to a secondary efﬂuent prior to

sand ﬁltration. It is desired to dose at 2.4 mg/L. Assuming an efﬂuent ﬂow of 6550 gpm, at what

rate (gallons per minute) should the polymer feed pump be set?

Polymer feed rate GPM:

gal

min

lbs polymer 100lbs polymer solution

gal polymer solution

2.4

6550∗8.34∗ ₁₀

∗

= 0.35

8.34lbs polymer solution

min

4.5lbspolymer

470

Chapter 22. Water Math

13. Polymer is being added at 0.3 mg/l in order to achieve a 92% capture efﬁciency for a belt press.

The feed to the belt press is 100 gallons per minute, containing 2.5% solids. Given the polymer

costs $460 per gallon of 4% active polymer with a speciﬁc gravity of 1.1. What is the cost of

polymer per dry ton of solids captured (10 points)

Solution:

lbs polymer required:

sludge

lbs sludge

0.3lbs polymer

100∗1440^gal

∗8.34

∗

day

gal sludge 1,000,000lbs sludge

polymer

day

= 0.36^lbs

gallons polymer solution required:

polmyer

day

polymer solution

day

8.34∗1.1lbs polymer solution

lbs polymer

polymer solution

0.36^lbs

= x^gal

∗

∗ 0.04_lb

gal polymer solution

polymer solution

day

=0.982^gal

Polymer cost:

0.982gal polymer soultion

$460

∗

gallon polymer soultion

day

$451.26

day

Dry tons of solids captured:

sludge

0.92 lbs solids captured

100∗1440^gal

∗

8.34∗0.025 lbs solids

ton solids

2000lbs solids

day

gal sludge

lbs solids

13.81tons dry solids

day

Polymer cost per dry ton of solids captured:

$451.26perday

= $32.67

13.81tons dry solids per day

14. A ﬂow of 5 MGD is being treated with 9.8 mg/l aluminum using liquid alum of 48% strength and

SG of 1.32. Alum has 19% aluminum. If the liquid alum costs $1.62 per gallon, what is the cost per

day

Solution:

lbs aluminum required:

aluminum

day

aluminum

= 408.7^lbs

day

5MGD∗8.34∗9.8^lbs

Alum needed to meet this dosing need:

liquid alum

day

lbs liquid alum

gal liquid alum

408.7^lbs

0.19^lbs

= x^gal

∗ 8.34 ∗ 1.32 _per

= 407^gal

∗ 0.48_lb

∗

aluminum

day

lbs alum

liquid alum

aluminum

lb alum

liquid alum

day

liquid alum

day

=⇒ x^gal

408.7

8.34∗1.32∗0.48∗0.19

liquid alum

day

Cost per day=407^gal

∗

= $659.45

gal liquid alum

$1.62

15. Prior to sand ﬁltration, a secondary efﬂuent ﬂow of 5 MGD is dosed with 0.75% strength poly-

22.27 Dewatering math problems

471

mer solution to achieve a dose of 1.5mg/l of polymer. a) What is the lbs of dry polymer per day

necessary to treat this efﬂuent, and b) What is the required GPM feed of the 0.75% polymer:

Solution:

Correct Answer:

lbs polymer

a) lbs of dry polymer required (lbs formula)=5MGD∗8.34∗1.5 = 62.55

day

gal

min

day

∗1440

polymer

day

min

b) Flow rate of 0.75% strength polymer = 62.55^lbs

∗8.34∗7500

gal

1,000,000

gal

62.55∗1,000,000

= 0.7GPM −Correct Answer

1440∗8.34∗7,500

=⇒ x

min

Incorrect Answer 1:

a) lbs of dry polymer required (lbs formula)=5MGD∗8.34∗1.5∗0.75 = 46.9

lbs polymer

day

gal

min

day

∗1440

polymer

day

min

b) Flow rate of 0.75% strength polymer = 62.55^lbs

∗8.34∗7500

gal

1,000,000

gal

min

46.9∗1,000,000

=⇒ x

= 0.5GPM −Incorrect Answer#1

1440∗8.34∗7,500

Incorrect Answer #2:

62.6GPM −Incorrect Answer#2

Incorrect Answer#3:

12.6GPM −Incorrect Answer#3

16. If a chemical costs $30 per ton, how much will it cost per year to treat a ﬂow of 15 MGD if the

average dose is 18 mg/L?

Solution:

Tons of chemical required per year: (use lbs formula)

days

lbs

ton

411 tons

year

15 MGD∗18

∗8.34 _day∗365 _{year 2000lbs}

∗

Chemical cost:

411 tons $30

∗

= $12,328 per year

ton

year

17. Prior to sand ﬁltration, a secondary efﬂuent ﬂow of 5 MGD is dosed with 0.75% strength polymer

solution to achieve a dose of 1.5mg/l of polymer. What is the required GPM feed of the 0.75

a. 12.6 GPM

b. 62.6 GPM

*c. 0.7 GPM

d. 0.5 GPM

Practice Problems - Pumping Power Requirements

1. If a pump is operating at 2,200 gpm and 60 feet of head, what is the water horsepower? If the pump

efﬁciency is 71%, what is the brake horsepower?

2. The water horsepower of a pump is 10Hp and the brake horsepower output of the motor is 15.4Hp

What is the efﬁciency of the pump?

3. The water horsepower of a pump is 25Hp and the brake horsepower output of the motor is 48Hp

What is the efﬁciency of the pump?

4. The efﬁciency of a well pump is determined to be 75%. The efﬁciency of the motor is estimated at

472

Chapter 22. Water Math

94%. What is the efﬁciency of the well?

5. If a motor is 85% efﬁcient and the output of the motor is determined to be 10 BHp, what is the

electrical horsepower requirement of the motor?

6. The water horsepower of a well with a submersible pump has been calculated at 8.2 WHp. The

Output of the electric motor is measured as 10.3BHp. What is the efﬁciency of the pump?

7. Water is being pumped from a reservoir to a storage tank on a hill. The elevation difference be-

tween water levels is 1200 feet. Find the pump size required to ﬁll the tank at a rate of 120 gpm.

Express your answer in horsepower.

25hp pump is used to dewater a lake. If the pump runs for 8 hours a day for 7 days a week, how

much will it cost to run the pump for one week? Assume energy costs $0.07 per kilowatt hour.

9. A pump station is used to lift water 50 feet above the pump station to a storage tank. The pump rate

is 500gpm. If the pump has an efﬁciency of 85% and the motor has an efﬁciency of 90%, ﬁnd each

of the following: Water Horsepower, Brake Horsepower, Motor Horsepower, and Wire-to-Water

Efﬁciency.

10. Find the brake horsepower for a pump given the following information: Total Dynamic Head = 75

feet, Pump Rate = 150 gpm, Pump Efﬁciency = 90%, Motor Efﬁciency = 85%

11. Water is being pumped from a reservoir to a storage tank on a hill. The elevation difference be-

tween water levels is 1200 feet. Find the pump size required to ﬁll the tank at a rate of 120 gpm.

Express your answer in horsepower.

Solutions:

1. Solution:

water Hp = ﬂow * head

2,200GPM ∗60 ft ∗

= Water Hp = 33.3Hp

3,960GPM − ft

pump Hp = brake Hp * pump efﬁciency

33.3

0.71

brake Hp =

= Brake Hp = 47Hp

2. Solution:

10BHp

η_p=

×100 = 65%

15.4EHp

3. Solution:

25 Water Hp

48 brake Hp

η_p=

×100 = 52%

22.27 Dewatering math problems

473

4. Solution:

Well e f ficiency = η_m∗η_p=⇒ 0.94×0.75 = 0.705×100 = 71%

5. Solution:

10BHp

= 12EHp or Input Hp

0.85

6. Solution:

8.2 W Hp

10.3 BHp

η_p=

×100 = 80%

7. Solution:

water Hp = ﬂow * head

Water Hp = 120 gpm∗1,200 ft ∗

= 37 Hp

3,960 gpm−ft

8. Solution:

0.746 kW 8 hrs 7 days $0.07

$73.1

week

25 Hp

∗

day

month kWh

9. Solution:

water Hp = ﬂow * head

Water Hp = 500 gpm∗50 ft ∗

= 6.3 WHp

3,960 gpm−ft

water Hp

Pump efﬁciency =

brake Hp

pump Hp

Pump efﬁciency

=⇒ brake Hp =

6.3

0.85

brake Hp =

= 7.4 Hp

brake Hp

input Hp

brake Hp

motor efﬁciency 0.9

7.4

Motor efﬁciency =

=⇒ input Hp =

= 8.2 Hp

474

Chapter 22. Water Math

Wire−to−water efﬁciency = η_m∗η_p=⇒ 0.9×0.85×100 = 77%

10. Solution:

water Hp = ﬂow * head

3,960GPM − ft

150 GPM∗75ft∗

= Water Hp = 2.8Hp

pump Hp = brake Hp * pump efﬁciency

2.8

0.9

brake Hp =

= Brake Hp = 3.1Hp

23. Wastewater Treatment - Future

23.1 Water Resource Recovery Facility - WRRF

•

Originally, the function of a wastewater treatment plant was collection, treatment and disposal

which was driven solely by the need to reduce human disease and to protect the environment.

It was soon realized that wastewater contains valuable elements like organic matter, phosphorus,

nitrogen, rare metals and thermal energy

The scope of wastewater treatment has now evolved to encompass recovering valuable resources

contained in the wastewater.

• The wastewater treatment plant is now known as a water resource recovery facility (WRRF).

•

This change reﬂects the new focus on the products and beneﬁts of treatment rather than its original

and only objective - treating water for sanitation.

•

The WRRF is being adapted into the concept of the circular economy in which products, mate-

rials (and raw materials) remain in the economy for as long as possible, and waste is treated as

secondary raw materials that can be recycled to process and re-used. This distinguishes it from a

linear economy which is based on the: "take-make-use-dispose" system, in which waste is usually

the last stage of the product life cycle.

23.1.1 Water

•

Municipal wastewater reuse offers the potential to signiﬁcantly increase the nation’s total available

water resources. Of the 32 billion gallons of treated wastewater discharged nationally, approxi-

mately 12 billion gallons of treated wastewater is discharged each day to an ocean or estuary.

WRRF is key to the use of treated wastewater, or “reclaimed” water, for beneﬁcial purposes such

as drinking, irrigation, or industrial uses—is one option that has helped some communities signiﬁ-

cantly expand their water supplies.

•

Wastewater can be treated to various qualities to satisfy demand from different sectors, including

industry and agriculture. It can be used to maintain the environmental ﬂow, or even reused as

476

Chapter 23. Wastewater Treatment - Future

drinking water.

•

Wastewater treatment is one solution to the water scarcity issue, and also to the problem of water

security, freeing water resources for other uses or for preservation.

23.1.2 Energy

•

In the US, municipal wastewater treatment plants consume 30 terawatt-hours per year of electricity

which is about 0.1% of the total electrical consumption.

•

WRRFs have the potential to be energy neutral or even net energy producers through comprehen-

sive energy management approaches, incorporating efﬁcient practices, and generating renewable

energy from their by-products, such as biosolids.

•

By making the treatment processes more energy efﬁcient and through the recovery of chemical or

caloriﬁc energy in the wastewater play a key role in reducing the carbon foot print of the WRRF.

Given the amount of chemical or caloriﬁc energy contained in wastewater, the goal is for the WR-

RFs to be energy neutral or even energy positive - which is to produce same or more energy than

the energy needed for treatment.

23.1.3 Nutrients

•

Nutrient recovery is the practice of recovering nutrients such as nitrogen and phosphorus from used

water streams that would otherwise be discarded/disposed and converting them into fertilizer used

for ecological and agricultural purposes.

•

Phosphate used in fertilizers is manufactured from phosphate rock using a very environmentally

detrimental mining process. Recovery/reuse of phosphate from wastewater mitigates dependence

on this mined mineral.

Nutrient recovery at a WRRF is accomplished using one of following two methods:

1. By precipitating phosphorous as struvite crystals using a dedicated reactor. The struvite crystals

are collected and resold as fertilizer which has a resale value ranging from $100-$600 per dry ton.

Beneﬁts of this method include:

• Generation of revenue from the fertilizer produced

•

Reduces fouling of equipment due to precipitation of struvite formed during the solids treat-

ment process

• Allows for meeting NPDES nutrients discharge limits

2. Nutrient recovery can also be achieved through the land application of biosolids. Beneﬁts of land

application of biosolids include:

•

Provide primary nutrients - nitrogen and phosphorous and secondary nutrients such as cal-

cium, iron, magnesium and zinc for crops.

• The organic carbon and organic matter in the biosolids help build better soils.

• Allows for sequestering carbon in the soil.

23.2 Challenges to WRRF

23.2.1 Constituents of Emerging Concern

Constituents of Emerging Concern (CECs) include a variety of substances such as medicines,

•

personal care products, ﬂame retardants, algal toxins, micorplastics, and many others that are not

currently federally regulated but known to occur in water.

23.2 Challenges to WRRF

477

•

Some of these CECs, incuding hormones, PFAS, and endocrine disruptors are known to pose health

risks to humans and aquatic life.

Although some CECs that reach WRRFs are destroyed through wastewater treatment and solids

processing, some recalcitrant microconstituents and their metabolites may pass through the treat-

ment process intact and may end up in the efﬂuent or biosolids.

•

Both, the dose (concentration) of the CEC present and frequency/duration of exposure is important

for interpreting possible risk to ecological and human health.

The CECs move in their complex cycling through surface and groundwaters across the planet -

from potable water to wastewater and viceversa as potable water is converted into wastewater fol-

lowed by uptake of the treated wastewater by the potable water supply system through groundwater

contaminated by the percolation of these CECs in land applied wastewater biosolids or through the

CECs in the treated wastewater discharge to surface water such as a lake or river.

The CECs concentrations in the plant inﬂuent range typically in nano-g/L to micro-g/L , in efﬂuent

from non-detect to nano-g/L, and in biosolids the concentrations vary from micro-g/kg to mg/kg.

•

23.2.2 Decentralized and Distributed Systems

•

To make the wastewater treatment systems more sustainable and to overcome the issues of a cen-

tralized system where wastewater is collected from various areas and cities in urban areas and

conveyed to a centrally located plant for treatment, it is imperative to consider

Distributed systems are in different geographical locations, but are linked to a central system ei-

ther physically, or by management. Decentralized systems can be located in a different geograph-

ical location, but are not linked physically, or are not managed under the umbrella of a centralized

system.

•

Through the selection of correct locations and appropriate technologies, distributed or decentralized

systems can:

– Provide environmental beneﬁts, such as nutrient and pathogen removal

–

Provide water for direct potable reuse and non-potable water in both rural and urban settings

for purposes such as ﬂushing, cooling and heating, landscaping, and subsurface irrigation

drip.

23.2.3 Climate Change

•

Climate change has directly impacted water resources by altering precipitation patterns, severe

drought and ﬂoods, snowpack amount, elevation, stream ﬂow, and rising sea levels.

This has created a direct need for utilities to manage local water resources to lessen the potential

impact of climate change.

By increasing water reuse, developing resiliency and other actions, WRRFs can be a leader in

ﬁghting and preparing for climate change effects.

Driven largely by climate change factors, lower carbon footprint - the amount of carbon dioxide

and other carbon compounds emitted due to the consumption of fossil fuels (energy), and lower

energy demands are now being factored in when assessing wastewater treatment options.

478

Chapter 23. Wastewater Treatment - Future

Source: Waste to Resource - World Bank

24. Wastewater Treatment - Careers

•

A career in wastewater treatment provides prospects of a stable and well paid employment with

added perk of protecting the environment and the health of the community.

Wastewater treatment plants are facing challenges to ﬁll the positions created by the retirement of

the baby boomers. It is estimated 30-50% of the current employees will retire within the next 10

years.

•

There are a variety of career paths which require different skill sets and training. From high school

graduates to PhDs.

Although actual renumeration and beneﬁts depend on the size and location of the plant, in general

wastewater jobs offer above average wages

Partial List of Wastewater Career Positions

Plant Operators

Collections Workers

Construction Inspectors

Civil Engineers

Electrical Maintenance Worker

Laboratory Technician

Mechanical Mntnc. Worker

Environmental Specialists

Electrical Engineers

Instrumentation Mntnc. Worker

Warehouse Workers

Surveyors

Mechanical Engineers

Ofﬁce Assistants

Public Info. Specialists

Contracts Administrators

Health and Safety Specialist

Planners and Schedulers

Finance and Accounting

Buyer

CAD and Graphics Designers

25. Glossary

ACID : (1) A substance that tends to lose a proton. (2) A substance that dissolves in water with the forma-

tion of hydrogen ions. (3) A substance containing hydrogen which may be replaced with metals to form

salts. (4) A substance that is corrosive. (5) A substance that may lower pH

ACIDITY : (A) Measure of the ability to neutralize alkaline (hi pH) substances. (B) The capacity of

water or wastewater to neutralize bases. Acidity is expressed in milligrams per liter of equivalent calcium

carbonate.

ACRE FOOT : A volume of water one (1) foot deep and one (1) acre in area, or 43,560 cubic feet.

ACTIVATED CARBON : Form of carbon processed to have small, low–volume pores that increase the

surface area available for adsorption or chemical reactions.

ACTIVATED SLUDGE : (A) A biological water treatment technology commonly used in municipal

wastewater treatment systems. Sometimes private industry will harness this technique to reduce cer-

tain pollutants, such as BOD and COD (see deﬁnitions below), but usually only due to compliance con-

cerns. (B) Sludge withdrawn from the secondary clariﬁer in the activated sludge process, consisting of

micro–organisms, non–living organic matter, and inorganic materials.

ACTIVATED SLUDGE PROCESS : A common method of disposing of pollutants in biological wastewa-

ters. In the process, large quantities of air are bubbled through wastewaters that contain dissolved organic

substances in open aeration tanks. Bacteria and other types of microorganisms present in the system need

oxygen to live, grown, and multiply in order to consume the dissolved organic "food" or pollutants in the

waste. After several hours in a large holding tank, the water is separated from the sludge of bacteria and

discharged from the system. Most of the activated sludge is returned to the treatment process, while the

remainder is disposed of by one of several acceptable methods.

ACTUATOR : Device used to operate a valve using electric, pneumatic or hydraulic means. Often used for

remote control or sequencing of valve operations.

ADAPTER SPOOL : An extension which is added to a short face–to–face valve, to conform to standard

482

Chapter 25. Glossary

API 6D face–to–face dimensions.

ADVANCED WASTE TREATMENT : (A) A treatment technology used to produce an extremely high–quality

discharge. (B) Any process of water renovation that upgrades treated wastewater to meet speciﬁc reuse

requirements. Typical processes include chemical treatment and pressure ﬁltration. Also called tertiary

treatment.

AERATION : (A) The process of bringing about intimate contact between air and a liquid. (B) The pro-

cess of adding air to wastewater to provide dissolved oxygen for aerobic bacterial treatment, to freshen

wastewater and to keep solids in suspension.

AERATION TANK : A chamber for injecting air into water.

AEROBES : Bacteria that must have molecular (dissolved) oxygen (DO) to survive.

AEROBIC : (A) A condition in which atmospheric or dissolved molecular oxygen is present in the aquatic

(water) environment. (B) requiring free oxygen for respiration. Refers to types of bacteria commonly

found in water and wastewater treatment systems.

AEROBIC BACTERIA : Bacteria which will live and reproduce only in an environment containing oxy-

gen which is available for their respiration (breathing), namely atmospheric oxygen or oxygen dissolved

in water. Oxygen combined chemically, such as water molecules (H2O), cannot be used for respiration by

aerobic bacteria.

AIR END : A term referring to the side (or parts) of the pump that come into contact with shop/compressed

air or natural gas. This applies to any air operated pump including Air Operated Diaphragm Pumps, air

operated piston pumps and air operated drum pumps.

AIR LIFT : A type of pump. This device consists of a vertical riser pipe in the wastewater or sludge to be

pumped. Compressed air is injected into a tall piece at the bottom of the pipe. Fine air bubbles mix with

the wastewater or sludge to form a mixture lighter than the surrounding water which causes the mixture

to rise in the discharge pipe to the outlet. An air–lift pump works like the center of a stand in a percolator

coffee pot.

AIR TEST : A method of inspecting a sewer pipe for leaks. Inﬂatable or similar plugs are placed in the

line, and the space between these plugs is pressurized with air. A drop in pressure indicates the line or run

being tested has leaks.

AIR: operated ejectors or centrifugal pumps.

ALGAE : A class of microscopic plant life that contain chlorophyll, live ﬂoating (suspended) in water

or are attached to rocks, walls and other surfaces, and grow and multiply through photosynthesis. Algae

produce oxygen during sunlight hours, use oxygen during darkness and affect the pH and DO levels in

water.

ALGAL BLOOM : Sudden, massive growths of algae that develop in lagoons, lakes and reservoirs.

ALIQUOT : Portion of a sample. Often an equally divided portion of a sample.

ALKALINITY : The capacity of water to neutralize acids, a property imparted by the water’s content of

carbonates, bicarbonates, hydroxides, and occasionally borates, silicates, and phosphates. Alkaline ﬂuids

have a pH value over 7.

ALL WELDED CONSTRUCTION : Pertains to a valve construction in which the body is completely

welded and cannot be disassembled and repaired in the ﬁeld.

483

ANAEROBIC : A biological environment that is deﬁcient in all forms of oxygen, especially molecular

oxygen, nitrates and nitrites. The decomposition by microorganisms of waste organic matter in wastewater

in the absence of dissolved oxygen is classed as anaerobic.

ANAEROBIC BACTERIA : Bacteria that live and reproduce in an environment containing no “free” or

dissolved oxygen. Anaerobic bacteria obtain their oxygen supply by breaking down chemical compounds

which contain oxygen, such as sulfate (SO 2–).

ANAEROBIC DECOMPOSITION : The decay or breaking down of organic material in an environment

containing no “free” or dissolved oxygen.

ANAEROBIC DIGESTION : Anaerobic bacteria (saprophytic and methane fermenters) decompose

wastewater solids (complex organic material) in two steps into – 1) volatile acids, and 2) methane gas,

carbon dioxide and water in the absence of dissolved oxygen. Specially designed basins, digesters, are

used to carry out the digestion processes, prevent air from entering and to capture the methane gas. The

sludge layer at the bottom of lagoons provides for similar solids stabilization processes.

ANCHOR PIN : A pin welded onto the body of ball valves. This pin aligns the adapter plate and restrains

the plate and gear operator from moving while the valve is being operated.

ANGLE VALVE : A variation of the globe valve, in which the end connections are at right angles to each

other, rather than being inline.

ANIONIC FLOCCULANT : Negatively charged ﬂocculant. Used in water treatment to aid solid / liquid

separation

ANOXIC : A biological environment that is deﬁcient in molecular oxygen, but may contain chemically

bound oxygen, such as nitrates and nitrites.

ANOXIC : Description of an environment without oxygen. In wastewater treatment anoxic processes are

typically used for the removal of nitrogen from wastewater.

ANTISCALENT : Material used to control scale formation in water systems such as boiler or cooling

water systems.

AOD : AOD stands for Air Operated Diaphragm (pump). These types of pumps are powered by com-

pressed air or gas, making them ideal for hazardous applications such as petroleum based products and

other ﬂammable materials. With certain materials of construction, such as steel or conductive plastics,

they are easily converted into fully explosion proof (Ex–Proof) pumps. Additionally, they can pull a

suction lift and are submersible when installed properly. AOD’s can also handle slurries with solids con-

centrations up to 30 AODD : AODD stands for Air Operated Double Diaphragm (pump). These types

of pumps are powered by compressed air or gas, making them ideal for hazardous applications such as

petroleum based products and other ﬂammable materials. With certain materials of construction, such as

steel or conductive plastics, they are easily converted into fully explosion proof (Ex–Proof) pumps. Addi-

tionally, they can pull a suction lift and are submersible when installed properly. AOD’s can also handle

slurries with solids concentrations up to 30 AQUIFER : A natural underground layer of porous materials

usually capable of yielding a supply of water.

ARSENIC : A heavy metal commonly regulated by wastewater discharge permits, but not commonly

found in industrial wastewaters. Other heavy metals include – Cadmium (Cd), Chromium(Cr), Copper

(Cu), Lead (Pb), Nickel (Ni), and Zinc (Zn).

ASPHYXIATION : An extreme condition often resulting in death due to a lack of oxygen and excess

484

Chapter 25. Glossary

carbon dioxide in the blood from any cause.

AVAILABLE CHLORINE : The amount of chlorine available in compound chlorine sources compared

with that of elemental (liquid or gaseous) chlorine.

AVERAGE MONTHLY DISCHARGE LIMITATION : The highest allowable discharge over a calendar

month

AVERAGE WEEKLY DISCHARGE LIMITATION : The highest allowable discharge over a calendar

week.

B.R.V. : BODY RELIEF VALVE – A relief valve (optional) installed on ball valves used in liquid service

to provide for the relief of excess body pressure caused by thermal expansion.

BACK WASH : Part of water ﬁlter, ion exchange or softener cycle that lifts up media bed to release and

wash away dirt and other foulants.

BACKFILL : (1) Material used to full in a trench or excavation. (2) The act of ﬁlling a trench or excava-

tion, usually after a pipe or some type of structure has been placed in the trench or excavation.

BACKFILL COMPACTION : (1) Tamping, rolling or otherwise mechanically compressing material used

as backﬁll for a trench of excavation. Backﬁll is compressed to increase its density so that it will support

the weight of machinery or other loads after the material is in place excavation. (2) Compaction of a

backﬁll material can be expressed as a percentage of the maximum compatibility, density or load capacity

of the material being used.

BACKFLOW : A reverse ﬂow condition, created by a difference in water pressures, which causes water

to ﬂow back into the distribution pipes of a potable water supply from any source or sources other than an

intended source. Also see BACKSIPHONAGE.

BACKFLUSHING : A procedure used to wash settled waste matter off upstream to prevent odors from

developing after a main line stoppage has been cleared.

BACKSEAT : A Shoulder on the stem of a valve which seals against a mating surface inside the bonnet to

permit replacement, under pressure, of stem seals or packing.

BACKSIPHONAGE : A form of backﬂow caused by a negative or below atmospheric pressure within a

water system. Also see BACKFLOW.

BACTERIA : Bacteria are microscopic living organisms They are a group of universally distributed, rigid,

essentially unicellular, microscopic organisms lacking chlorophyll. They are characterized as spheroids,

rod–like, or curved entities, but occasionally appearing as sheets, chains, or branched ﬁlaments.

BAFFLE : A ﬂat board or plate, deﬂector, guide or similar device constructed or placed in ﬂowing water,

wastewater, or slurry systems to cause more uniform ﬂow velocities, to absorb energy, and to divert, guide,

or agitate liquids (water, chemical solutions, slurry).

BALL : The spherical closure element of a ball valve.

BALL CHECK : A ﬁtting with a small ball that seals against a seat preventing ﬂow in one direction and

allowing ﬂow in the other direction.

BALL VALVE : A valve using a spherical closure element (ball) which is rotated thru 90° to open and

close the valve.

BALLING : A method of hydraulically cleaning a sewer or storm drain by using the pressure of a water

head to create a high cleansing velocity of water around the ball. In normal operation, the ball is restrained

485

by a cable while water washes past the ball at high velocity. Special sewer cleaning balls have an outside

tread that causes them to spin or rotate, resulting in a “scrubbing” action of the ﬂowing water along the

pipe wall.

BAR RACK : A screen composed of parallel bars, either vertical or inclined, placed in a sewer or other

waterway to catch debris. The screenings may be raked from it.

BARREL : (1) The cylindrical part of a pipe that may have a bell on one end. (2) The cylindrical part of a

manhole between the cone at the top and the shelf at the bottom.

BASE : (1) A substance which takes up or accepts protons. (2) A substance which dissociates (separates)

in aqueous solution to yield hydroxyl ions (OH–). (3) A substance containing hydroxyl ions which reacts

with an acid to form a salt or which may react with metalsvto form precipitates. (4) A substance that may

raise pH.

BASE : A substance that takes up or accepts protons, dissociates in water to produce hydroxyl (OH–) ions,

reacts with metals and is corrosive.

BDV : BLOW DOWN VALVE – A small ball valve that is installed on the aboveground end of an ex-

tended drain line. This valve also serves to vent body cavity pressure in the "block and bleed" mode.

BEDDING : The prepared base or bottom of a trench or excavation on which a pipe or other underground

structure is supported.

BEDDING COMPACTION : (1) Tamping, rolling or otherwise mechanically compressing material used

as bedding for a pipe or other underground structure to a density that will support expected loads. (2)

Bedding compaction can be expressed as a percentage of the maximum load capacity of the bedding

material. (3) Bedding compaction also can be expressed in load capacity or pounds per square foot.

BEDDING GRADE : (1) In a gravity–ﬂow sewer system, pipe bedding is constructed and compacted to

the design grade of the pipe. This is usually expressed in a percentage. A 0.5 percent grade would be a

drop of one–half of foot per hundred feet of pipe. (2) Bedding grade for a gravity–ﬂow sewer pipe can

also be speciﬁed as elevation above mean sea level at speciﬁc points.

BELL : (1) In pipe ﬁtting, the enlarged female end of a pipe into which the male end ﬁts. (2) In plumbing,

the expanded female end of a wiped joint.

BELL: AND–SPIGOT JOINT – A form of joint used on pipes which have an enlarged diameter or bell

at one end, and a spigot at the other which ﬁts into and is laid in the bell. The joint is then made tight by

lead, cement, rubber O–ring, or other jointing compounds or materials.

BELLEVILLE SPRING : A spring resembling a dished washer, used in some ball valves to push the seats

against the ball.

BERM : The earthen dike that surrounds ponds, lagoons and containment areas for hazardous material.

BEST EFFICIENCY POINT (B.E.P.) : The point on a pump’s performance curve that corresponds to the

highest efﬁciency.

BEVEL GEAR OPERATOR : Device facilitating operation of a gate or globe valve by means of a set of

bevel gears having the axis of the pinion gear at right angles to that of the larger ring gear. The reduction

ratio of this gearsetdetermines the multiplication of torque achieved.

BHP : BHP is the actual amount of horsepower being consumed by the pump as measured on a pony

brake or dynamometer.

486

Chapter 25. Glossary

BIOCHEMICAL OXYGEN DEMAND (BOD) : A quantitative measure of the oxygen needed by bacteria

and micro–organisms for the biological oxidation of organic wastes in a unit volume of wastewater. BOD

is generally measured in milligrams per liter (mg/l) of oxygen consumed over a ﬁve day period. Although

complete biological decomposition of organic waste requires about 20 days, the ﬁve day BOD is about

two–thirds of the total oxygen required and, therefore, is a practical measure of waste concentration. In

waste treatment language, BOD is most frequently stated as the percentage removed during treatment, or

remaining after treatment.

BIOCIDE : Chemical substance designed for killing living organisms in water. Often characterized by

type of organism killed – bactericide, fungicide or algaecide.

BIOLOGICAL OXIDATION : The process by which bacteria and other types of microorganisms consume

dissolved oxygen and organic substances in biological wasterwater. The energy released is then used to

convert organic carbon into carbon dioxide and cellular material.

BIOMASS : Amass or clump of living organisms feeding on wastes in wastewater, dead organisms and

other debris. The mass may protect the organisms, as well as store food supplies. Also called ZOOGLEAL

MASS.

BIOSOLIDS : A primarily organic solid product, produced by wastewater treatment processes, that can be

beneﬁcially recycled. The word biosolids is replacing the word sludge.

BIOSOLIDS CAKE : Solid discharge from a dewatering apparatus.

BIT : (1) Cutting blade used in rodding (pipe cleaning) operations. (2) Cutting teeth on the auger head of a

sewer boring tool.

BLANK : A bottle containing only dilution water or distilled water, but the sample being tested is not

added. Tests are frequently run on a SAMPLE and a BLANK and the differences are compared.

BLOCK AND BLEED : The capability of obtaining a seal across the upstream and downstream seat rings

of a valve when the body pressure is bled off to atmosphere thru blow down valves or vent plugs. Useful

in testingfor integrityof seat seals and in accomplishing minor repairs under pressure.

BLOCKAGE : (1) Partial or complete interruption of ﬂow as a result of some obstruction in a sewer. (2)

When a collection system becomes plugged and the ﬂow backs up, “blockage.”

BLOW DOWN (BLEED: off) – terms to describe the deliberate rejection of water from a system such

as boiler or cooling water system. Typically done to control system’s water total dissolved system or

conductivity.

BLUE: GREEN ALGAE – Varieties of algae characterized by their bluish–green color. The appearance of

blue–green algae indicates unhealthy conditions in lagoon cells, often associated with organic overloading

and lack of adequate dissolved oxygen.

BOD : Biochemical Oxygen Demand. The rate at which organisms use the oxygen in water or wastewater

while stabilizing decomposable organic matter under aerobic conditions. In decomposition, organic matter

serves as food for the bacteria and energy results from its oxidation. BOD measurements are used as a

measure of the organic strength of wastes in water.

BODY : The principal pressure containing part of a valve, in which the closure element and seats are

located.

BOLTED BONNET : A bonnet which is connected to a valve body with bolts or studs and nuts.

487

BOLTED CONSTRUCTION : Describes a valve construction in which the pressure shell elements are

bolted together, and thus can be taken apart and repaired in the ﬁeld.

BONNET : The top part of a valve, attached to the body, which contains the packing gland, guides the

stem, and adapts to extensions or operators.

BORE (OR PORT) : The inside diameter of the smallest opening through a valve, e.g., inside diameter of

a seat ring, diameter of hole through ball in a ball valve.

BRANCH MANHOLE : A sewer or drain manhole which has more than one pipe feeding into it. A

standard manhole will have one outlet and one inlet. A branch manhole will have one outlet and two or

more inlets.

BRANCH SEWER : A sewer that receives wastewater from a relatively small area and discharges into a

main sewer servicing more than one branch sewer area.

BUBBLE: TIGHT SHUT–OFF – A phrase used in describing the sealing ability of a valve. During air

pressure testing of a new valve in the closed position, leakage past the seats is collected and bubbled thru

water. To qualify as "bubble tight," no bubbles should be observed in a prescribed time span.

BUCKET : (1) A special device designed to be pulled along a sewer for the removal of debris from the

sewer. The bucket has one end open with the opposite end having a set of jaws. When pulled from the jaw

end, the jaws are automatically opened. When pulled from the other end, the jaws close. In operation, the

bucket is pulled into the debris from the jaw end and to a point where some of the debris has been forced

into the bucket. The bucket is then pulled out of the sewer from the other end, causing the jaws to close

and retain the debris. Once removed from the manhole, the bucket is emptied and the process repeated.

(2) A conventional pail or bucket used in BUCKETING OUT and also for lowering and raising tools and

materials from manholes and excavations.

BUCKET BAIL : The pulling handle on a bucket machine.

BUCKET MACHINE : A powered winch machine designed for operation over a manhole. The machine

controls the travel of buckets used to clean sewers.

BUCKETING OUT : An expression used to describe removal of debris from a manhole with a pail on

a rope. In balling or high–velocity cleaning of sewers, debris is washed into the downstream manhole.

Removal of this debris by scooping it into pails and hauling debris out is called “bucketing out.”

BUFFER : A solution or liquid whose chemical makeup neutralizes acids or bases without a great change

in pH.

BULKING : Clouds of billowing sludge that occur throughout secondary clariﬁers and sludge thickeners

when the sludge does not settle properly. In the activated sludge process bulking is usually caused by

ﬁlamentous bacteria or bound water.

BULKING SLUDGE : A phenomenon that occurs in activated sludge plants whereby the sludge occupies

excessive volumes and will not concentrate readily. This condition refers to a decrease in the ability of

the sludge to settle and consequent loss over the settling tank weir. Bulking in activated sludge aeration

tanks is caused mainly by excess suspended solids (SS) content. Sludge bulking in the ﬁnal settling tank

of an activated sludge plant may be caused by improper balance of the BOD load, SS concentration in the

mixed liquor, or the amount of air used in aeration.

BURIED SERVICE : An application in which valves are installed in lines which are buried below ground

level.

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Chapter 25. Glossary

BUTT WELD END (BWE) : The end connection of a valve suitably prepared for butt welding to a con-

necting pipe.

BUTTERFLY VALVE : A short face–to–face valve which has a movable vane, in the center of the ﬂow

stream, which rotates 90 degrees as the butterﬂy valve opens and closes.

BVR : BALL VALVE REGULATOR – An automatic throttling valve controlling ﬂow or pressure in a

pipeline; comprising a package involving al ball valve actuator, positioner, and controlling instrument.

BYPASS : A system of pipes and valves permitting the diversion of ﬂow or pressure around a line valve.

BYPASS : A pipe, valve, gate, weir, trench or other device designed to permit all or part of a wastewater

ﬂow to be diverted from usual channels or ﬂow. Sometimes refers to a special line which carries the ﬂow

around a facility or device that needs maintenance or repair.

BYPASSING : The act of causing all or part of a ﬂow to be diverted from its usual channels. In a wastewa-

ter treatment plant, overload ﬂows should be bypassed into a holding pond for future treatment. 224

CADMIUM : A heavy metal commonly regulated by wastewater discharge permits and typically found

in the metal ﬁnishing industry. Other heavy metals include – Arsenic (As), Chromium(Cr), Copper (Cu),

Lead (Pb), Nickel (Ni), and Zinc (Zn).

CAKE SOLID DISCHARGE RATE : The dry solids cake discharge from a centrifuge, which is expressed

as: dry cake solids discharge rate = (dry solids feed rate) x (solids recovery).

CALCIUM AND MAGNESIUM SOAPS, MINERAL OILS, AND CERTAIN OTHER NON: fatty mate-

rial which tend to separate from water and coagulate as ﬂoatables or scums.

CARBON DIOXIDE : A common gas, CO2, found abundantly in air, is a product of bacterial respiration

and used by algae in photosynthesis. The concentration of carbon dioxide in the lagoon water governs the

pH of the lagoon.

CARCINOGEN : Any substance that tends to produce cancer in an organism.

CASING : The body of the pump which encloses the impeller. Primarily used in reference to centrifugal

pumps.

CAST : The form of a particular part of a valve, where the basic shape is formed by molding rather than

fabricating.

CASTING : A product or the act of producing a product made by pouring molten metal into a mold and

allowing it to solidify, thus taking the shape of the mold.

CATCH BASIN : A chamber or well used with storm or combined sewers as a means of removing grit

which might otherwise enter and be deposited in sewers.

CATIONIC FLOCCULANT : Positively charged high molecular weight polyelectrolyte water soluble

organic polymer designed to agglomerate solids in water substrates.

CAVITATION : The formation and collapse of a gas pocket or bubble on the blade of an impeller or gate

of a valve. The collapse of the bubble drives water into the impeller or gate with a terriﬁc force that can

cause pitting of the surface. Cavitation is indicated by loud hammering noises.

CENTRIFUGAL FORCE : A force associated with a rotating body. In the case of a pump, the rotating

impeller pushes ﬂuid on the back of the impeller blade, imparting motion. Since the motion is circular

there is a centrifugal force associated with it. The force pushes the ﬂuid against a ﬁxed pump casing

thereby pressurizing the ﬂuid and forcing it through the outlet.

489

CENTRIFUGAL PUMP : Centrifugal pumps are the most common type of pump in use today throughout

the world. A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the velocity

of a ﬂuid. Centrifugal pumps are commonly used to move liquids through a piping system. The ﬂuid

enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, ﬂowing

radially outward into a diffuser or volute chamber, from there it exits into the downstream piping system.

A centrifugal pump works by the conversion of the rotational kinetic energy, typically from an electric

motor or engine, to an increased static ﬂuid pressure. This action is described by Bernoulli’s principle.

The rotation of the pump impeller imparts kinetic energy to the ﬂuid as it is drawn in from the impeller

eye (center) and is forced outward through the impeller vanes to the periphery. As the ﬂuid exits the

impeller, the ﬂuid kinetic energy (velocity) is then converted to (static) pressure due to the change in area

the ﬂuid experiences in the volute section. Typically, the volute shape of the pump casing (increasing

in volume), or the diffuser vanes (which serve to slow the ﬂuid, converting to kinetic energy in to ﬂow)

are responsible for the energy conversion. The energy conversion results in an increased pressure on the

downstream side of the pump, causing ﬂow.

CENTRIFUGE : A mechanical device that uses centrifugal or rotational forces to separate solids from

liquids.

CHAIN WHEEL OPERATED VALVE : An overhead valve operated by a chain drive wheel instead of a

handwheel.

CHARACTERIZED GATE OR BALL : A ball or gate, the shape of whose port has been specially altered

to provide a speciﬁc throttling capability.

CHECK VALVE : A one–directional valve which is opened by the ﬂuid ﬂow in one direction and closed

automatically when the ﬂow stops or is reversed.

CHEMICAL GROUT : Two chemical solutions that form a solid when combined. Solidiﬁcation time is

controlled by the strength of the mixtures used and the temperature.

CHEMICAL OXYGEN DEMAND (COD) : A quantitative measure of the amount of oxygen required to

oxidize all organic compounds in a unit volume on wastewater – non–biodegradable as well as the BOD.

The COD level can be determined more readily than BOD, but this measurement does not indicate how

much of the waste can be decomposed by biological oxidation.

CHLORINATION : The application of chlorine to water, sewage, or industrial wastes, generally for

the purpose of disinfection, but frequently for accomplishing other chemical or biological wasterwater

treatment results.

CHLORINATOR : A metering device which is used to add chlorine to water.

CHLORINE CONTACT UNIT : A bafﬂed basin that provides sufﬁcient time for disinfection to occur.

CHLORINE DEMAND : Chlorine demand is the difference between the amount of chorine added to

wastewater and the amount of residual chlorine remaining after a given contact time. Chlorine demand

may change with dosage, time temperature, pH, and nature and amount of the impurities in the water.

CHLORINE REQUIREMENT : The amount of chlorine which is needed for a particular purpose. Some

reasons for adding chlorine are reducing the number of coliform bacteria (Most Probable Number), obtain-

ing a particular chlorine residual, or oxidizing some substance in the water. In each case a deﬁnite dosage

of chlorine will be necessary. This dosage is the chlorine requirement.

CHLORINE RESIDUAL : The amount of free chlorine remaining after meeting chlorine demand under

490

Chapter 25. Glossary

given conditions and is necessary to complete disinfection.

CHOPPER PUMP : A chopper pump is a centrifugal pump, which is equipped with a cutting system

to facilitate chopping/maceration of solids that are present in the pumped liquid. The main advantage

of this type of pump is that it prevents clogging of the pump itself and of the adjacent piping, as all the

solids and stringy materials are macerated by the chopping system. Chopper pumps exist in various

conﬁgurations, including submersible and dry–installed design and they are typically equipped with an

electric motor to run the impeller and to provide torque for the chopping system. Due to its high solids

handling capabilities, the chopper pump is often used for pumping sewage, sludge, manure slurries, and

other liquids that contain large or tough solids.

CHROMIUM : A heavy metal commonly regulated by wastewater discharge permits and found in met-

als–related industries and products (including stainless steel). It is typically regulated in two forms – total

chromium and hexavalent chromium. Other heavy metals include – Arsenic (As), Cadmium (Cd), Copper

(Cu), Lead (Pb), Nickel (Ni), and Zinc (Zn).

CITY GATE : CITY GATE STATION – The metering and pressure reducing station where gas is trans-

ferred from a high pressure cross–country transmission line to a low pressure distribution piping system

within a city.

CLAPPER : The hinged closure element of a swing check valve.

CLARIFICATION : Any process or processes used to reduce the concentration of suspended matter in a

liquid, such as quiescent settling or sedimentation. Lagoons provide clariﬁcation across the cells and in

quiescent zones in aerated systems, allowing solids to settle into a sludge layer

CLARIFIER : A large, circular or rectangular tank that separates solids from the waste stream by settling

or ﬂotation.

CLEAN IN PLACE (CIP) : Method of cleaning the interior surfaces of pipes, vessels, process equipment,

ﬁlters and associated ﬁttings, without disassembly.

CLEAN WATER ACT : Federal legislation passed in 1972 creating the Environmental Protection Agency,

requiring a nationwide system for controlling pollutant discharges and providing for construction and

regulation of publicly owned treatment works.

CLEANOUT : An opening (usually covered or capped) in a wastewater collection system used for insert-

ing tools, rods or snakes while cleaning a pipeline or clearing a stoppage.

CLOSURE ELEMENT : The moving part of a valve, positioned in the ﬂowstream which controls ﬂow

thru the valve. Ball. Gate, Plug, Clapper, Disc, etc., are speciﬁc names for closure elements.

CO: precipitation – Term to describe compound used in water treatment to aid precipitation of substances

normally soluble under the conditions employed. Common co–precipitants used in water are iron, alu-

minum, calcium and magnesium.

COAGULANTS : Chemicals that cause very ﬁne particles to clump (ﬂoc) together into larger particles.

This makes it easier to separate the solids from the water by settling, skimming, draining or ﬁltering.

COAGULATION : The agglomeration of colloidal or suspended matter brought about by the addition of

some chemical to the liquid, by contact, or by other means. Involves destabilization for repulsive electrical

charges to permit agglomeration of colloid particles in water. This process aids the clariﬁcation of water.

COLIFORM : A type of bacteria. The presence of coliform–group bacteria is an indication of possible

pathogenic bacterial contamination. The human intestinal tract is one of the main habitats of coliform bac-

491

teria. They may also be found in the intestinal tracts of warm–blooded animals, and in plants, soil, air, and

the aquatic environment. Fecal coliforms are those coliforms found in the feces of various warm–blooded

animals; whereas the term “coliform” also includes various other environmental sources.

COLIFORM : A type of bacteria. The presence of coliform–group bacteria is an indication of possible

pathogenic bacterial contamination. The human intestinal tract is one of the main habitats of coliform bac-

teria. They may also be found in the intestinal tracts of warm–blooded animals, and in plants, soil, air, and

the aquatic environment. Fecal coliforms are those coliforms found in the feces of various warm–blooded

animals; whereas the term “coliform” also includes various other environmental sources.

COLIFORM BACTERIA : Live in everyone’s intestinal track. They are considered non–pathogenic.

COLIFORM ORGANISMS : A group of bacteria recognized as indicators of fecal pollution (see also

escherichia coliform).

COLLECTION SYSTEM : A network of pipes, manholes, cleanouts, traps, siphons, lift stations and

other structures used to collect all wastewater and wastewater–carried wastes of an area and transport

them to a treatment plant or disposal system. The collection system includes land, wastewater lines and

appurtenances, pumping stations and general property.

COLORIMETRIC MEASUREMENT : A means of measuring unknown chemical concentrations in water

by MEASURING A SAMPLE’S COLOR INTENSITY. The speciﬁc color of the sample, developed by

addition of chemical reagents, is measured with a photoelectric colorimeter or is compared with “color

standards” using, or corresponding with, known concentrations of the chemical.

COLORIMETRIC MEASUREMENT : A means of measuring unknown chemical concentrations in water

by MEASURING A SAMPLE’S COLOR INTENSITY. The speciﬁc color of the sample, developed by

addition of chemical reagents, is measured with a photoelectric colorimeter or is compared with “color

standards” using, or corresponding with, known concentrations of the chemical.

COMBINED SEWER : Carries both sanitary sewage and storm water run–off.

COMMINUTOR : A device used to reduce the size of the solid chunks in wastewater by shredding (com-

minuting). The shredding action is like many scissors cutting or chopping to shreds all the large inﬂuent

solids material in the wastewater.

COMMUNITY WASTEWATER SYSTEM : A public wastewater system which has at least 15 service

connection or treats 5,000 gallons or more of wastewater per day. The term “community wastewater sys-

tem” is used only to identify the public wastewater systems which must be operated by certiﬁed operators.

COMPOSITE (PROPORTIONAL) SAMPLE : A collection of individual samples obtained at regular

intervals during a 24–hour period. Each individual sample is combined with the others in proportion to the

rate of ﬂow when the sample was collected. The resulting mixture, or composite, forms a representative

sample and is analyzed to determine the average conditions during the sampling period.

COMPUTED PER CAPITA CONTRIBUTION : The computed wastewater contribution from a domestic

area, based on the population of the area. In the United States, the daily average wastewater contribution is

considered to be 100 gallons per capita per day (100GPCD).

COMPUTED TOTAL CONTRIBUTION : The total anticipated load on a wastewater treatment plant or

the total anticipated ﬂow in any collection system area based on the combined computed contributions of

all connections to the system.

CONCENTRATE : The high TDS discharge from a reverse osmosis ﬁltration process.

492

Chapter 25. Glossary

CONCRETE CRADLE : A device made of concrete that is designed to support sewer pipe. 225

CONDENSATE : steam that has lost heat and condensed into water.

CONDUCTIVITY : Transmittance of an electric current through water. Usually measured in microsiemens

per centimeter (uS/cm) or micromho per centimeter (umho/cm).

CONFINED SPACE : Conﬁned space means a space that: A. Is large enough and so conﬁgured that an

employee can bodily enter and perform assigned work; and B. Has limited or restricted means for entry or

exit; and C. Is not designed for continuous employee occupancy.

CONTAMINATION : The introduction into water of microorganisms, chemicals, toxic substances, wastes,

or wastewater in a concentration that makes the water unﬁt for its next intended use.

CONTROL VALVE : A valve that controls a process variable, such as pressure, ﬂow or temperature by

modulating its opening in response to a signal from a controller.

CONTROL VOLUME : Limits imposed for the theoretical study of a system. The limits are usually set to

intersect the system at locations where conditions are known.

CONTROLLER : A device that measures a controlled variable, compares it with a predetermined setting

and signals the actuator to read just the opening of the valve in order to re–establish the original control

setting.

COPPER : A heavy metal commonly regulated by wastewater discharge permits. It is found in the

metal ﬁnishing and electrical industries. Other heavy metals include – Arsenic (As), Cadmium (Cd),

Chromium(Cr), Lead (Pb), Nickel (Ni), and Zinc (Zn).

CORROSION : The gradual decomposition or destruction of a material due to chemical action, often due

to an electrochemical reaction. Corrosion starts at the surface of a material and moves inward, such as the

chemical action upon manholes and sewer pipe materials.

CORROSION INHIBITOR : chemical additive designed to control / minimize metal corrosion in water

system.

COULISSE : Of or using runners or slides as a guiding mechanism; as in a "Coulisse" style gate valve.

COUPLING : (1) A threaded sleeve used to connect two pipes. (2) A device used to connect two adjacent

parts, such as pipe coupling, hose coupling or drive coupling.

COUPON : A steel specimen inserted into wastewater to measure the corrosiveness of the wastewater.

The rate of corrosion is measured as the loss of weight of the coupon or change in its physical charac-

teristics. Measure the weight loss (in milligrams) per surface area (in square decimeters) exposed to the

wastewater per day.

CREST : The bottom edge of a weir plate.

CROSS CONNECTION : A connection between a drinking (potable) water system and an unapproved

water supply. For example, if you have a pump moving nonpotable water and hook into the drinking water

system to supply water for the pump seal, a cross connection or mixing between the two water systems

can occur. This mixing may lead to contamination of the drinking water.

CROSS CONNECTION : A connection between a drinking (potable) water system and an unapproved

water supply. For example, if you have a pump moving nonpotable water and hook into the drinking water

system to supply water for the pump seal, a cross connection or mixing between the two water systems

can occur. This mixing may lead to contamination of the drinking water.

493

CROSS CONNECTION : A connection between a storm drain system and a sanitary collection system.

CROSS: CONNECTION – A connection between a drinking water system and an unapproved system.

CRUSTACEANS : A class of microscopic water animals that consume large quantities of bacteria and

algae.

CRYOGENIC VALVE : A valve capable of functioning at cryogenic temperatures.

CYANIDE : A toxic element often found in wastewater from metal ﬁnishing industries. It’s commonly

regulated by wastewater permits.

CYCLE : A single complete operation or process returning to the starting point. A valve, stroked from full

open to full close and back to full open, has undergone one cycle.

CYCLES OF CONCENTRATION : Ratio of boiler or cooling water to make up (feed water). Typically

measured by monitoring total dissolved solids, conductivity, silica or chlorides.

CYLINDER OPERATOR : A power–piston valve operator using either hydraulic or pneumatic pressure.

A sealed piston converts applied pressure into a linear piston rod (stem) motion.

DAILY DISHARGE : The discharge of a pollutant measured during a calendar day or any 24 – hour

period that reasonably represents a calendar day for the purposes of sampling.

DAILY MAXIUM DISCHARGE : The highest allowable value for a daily discharge.

DAPHNIA : A crustacean commonly found in wastewater lagoons.

DATUM PLANE : A reference plane. A conveniently accessible known surface from which all vertical

measurements are taken or referred to.

DEADEND MANHOLE : A manhole located at the upstream end of a sewer and having no inlet pipe.

DEBRIS : Any material in wastewater found ﬂoating, suspended, settled, or moving along the bottom of

a sewer. This material may cause stoppages by getting hung up on roots or settling out in a sewer. Debris

includes grit, paper, rubber, silt, and all materials except liquid.

DECHLORINATION : The removal of chlorine from the efﬂuent of a treatment plant.

DEMULSIFIER (EMULSION BREAKER) : Chemical additive that destroys the emulsifying characteris-

tics water. Typically used separate stabilized (emulsiﬁed) oil in water.

DENITRIFICATION : A biological process by which nitrate is converted to nitrogen gas.

DENITRIFICATION : (1) The anoxic biological reduction of nitrate nitrogen to nitrogen gas. (2) The

removal of some nitrogen from a system. (3) An anoxic process that occurs when nitrite or nitrate ions are

reduced to nitrogen gas and nitrogen bubbles are formed as a result of this process.

DENITRIFICATION : (1) The anoxic biological reduction of nitrate nitrogen to nitrogen gas. (2) The

removal of some nitrogen from a system. (3) An anoxic process that occurs when nitrite or nitrate ions are

reduced to nitrogen gas and nitrogen bubbles are formed as a result of this process.

DENITRIFICATION : An anaerobic process that occurs when nitrite and nitrate ions are reduced to

nitrogen gas and bubbles are formed. These bubbles attach to sludge ﬂocs, causing rising sludge that ﬂoats

to the surface of secondary clariﬁers.

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Chapter 25. Glossary

DENITRIFICATION : Wastewater treatment process involved in the biological removal of nitrogen in

which the nitrite (NO2) is converted to nitrogen gas (N2)

DEQ : Department of Environmental Quality.

DETENTION TIME : The theoretical time that water may stay in a basin such as lagoon. It is the total

volume of the lagoon divided by the ﬂow rate. Usually expressed in days of time or in hours. The theoreti-

cal time water remains in a tank at a given discharge. The time required to ﬁll a tank at a given ﬂow or the

theoretical time required for a given ﬂow of wastewater to pass through a tank.

DETRITUS : The heavy, coarse mixture of grit and organic material carried by wastewater. (also called

grit).

DEWATER : To drain or remove water from an enclosure. A structure may be dewatered so that it can be

inspected or repaired. Dewater also means draining or removing water from sludge to increase the solids

concentration. 226

DEWATERING : Removing free water from a sludge or slurry to form a high solids cake. belt ﬁlter

presses, centrifuges, rotary fan presses and vacuum presses are dewatering devices.

DEWATERING PUMP : Any pump capable of removing water from an unwanted area. They are usually

small, portable pumps that run on single phase power, compressed air or a small engine, but can be large

permanently installed units as well.

DIAPHRAGM : A round, thin ﬂexible sealing device secured and sealed around its outer edge – and

sometimes around a central hole in the diaphragm – with its unsupported area free to move by ﬂexing.

DICHARGE MONITORING REPORT (DMR) : The monthly report required by the treatment plant’s

NPDES / OPDES discharge permit.

DIFFUSED: AIR AERATION – A diffused air activated sludge plant takes air, compresses it, and then

discharges the air below the water surface of the aerator through some type of air diffusion device.

DIFFUSER : A device used to break the air stream from the blower system into ﬁne bubbles in an aeration

tank or reactor.

DIGESTER : A tank in which sludge is placed to allow decomposition by microorganisms. Digestion may

occur under anaerobic (more common) or aerobic conditions.

DIGESTION : The biochemical decomposition of organic matter that results in the formation of mineral

and simpler organic compounds.

DIP : A point in the sewer pipe where a drain grade defect results in a puddle of standing water when

there is no ﬂow.

DIP TUBE : Extending the blow down valve on large gate valves requires a tube which is located inside

of the valve. The tube is called the "dip tube" and extends through the bonnet to the bottom of the body

cavity.

DISC : The closure element of a globe angle or small regulator valve. The disc (sometimes referred to

as "valve," "poppet" or "plug") moves to and from the seat in a direction perpendicular to the seat face.

Depends on stem force for tight shutoff.

DISCHARGE STATIC HEAD : The difference in elevation between the liquid level of the discharge tank

and the centerline of the pump. This head also includes any additional head that may be present at the

discharge tank ﬂuid surface.

495

DISINFECTION : The process designed to kill most microorganisms in water, including the destruction

or inactivation of pathogenic bacteria. Disinfection differs from sterilization which destroys all living

forms.

DISINFECTION : The process designed to kill or inactivate most microorganisms in wastewater, in-

cluding essentially all pathogenic (disease–casing) bacteria. There are several ways to disinfect, with

chlorination being the most frequently used in water and wastewater treatment plants.

DISINFECTION : The process designed to kill or inactivate most microorganisms in wastewater, in-

cluding essentially all pathogenic (disease–casing) bacteria. There are several ways to disinfect, with

chlorination being the most frequently used in water and wastewater treatment plants.

DISINFECTION : The process designed to kill or inactivate most microorganisms in water, including

essentially all pathogenic (disease–causing) bacteria. There are several ways to disinfect, with chlorine

being the most frequently used method in both water and wastewater systems.

DISPERSANT : A non–surface active compound or an active substance added to a suspension, usually a

mix, to increase the separation of particles and to prevent subsiding or clumping.

DISSOLVED AIR FLOTATION : A physical/chemical wastewater treatment technology that can be

cost–effectively used by industry to remove FOG (fats, oils and grease), suspended solids, and some

metals.

DISSOLVED AIR FLOTATION CLARIFIER (DAF) : a piece of equipment that used dissolved air to

ﬂoat suspended solids from water. Typically used when the suspended solids have a lower density than

water.

DISSOLVED OXYGEN (DO) : The oxygen dissolved in water, wastewater, or other liquid. DO is mea-

sured in milligrams per liter. If the DO of a sample of water is 2 mg/L, it means that there are 2lbs of

oxygen in 1 mil lb of water.

DISSOLVED SOLIDS : The salts and other residues left after evaporation of water that has been passed

through a laboratory ﬁlter. Dissolved solids cannot be ﬁltered out. Some colloidal solids may not be in

true solution, but if they pass through the standard membrane ﬁlter, they are considered dissolved solids.

(See suspended solids)

DIURNAL : Having a daily cycle; usually a 24–hour period from 12:00am to 12:00pm.

DO : Dissolved Oxygen – An indication of how much oxygen is present in water. If a facility discharges

directly to a stream or river, it will usually have a permit limit related to dissolved oxygen.

DRAGLINE : A machine that drags a bucket down the intended line of a trench to dig or excavate the

trench. Also used to dig holes and move soil or aggregate.

DRAIN PLUG : A ﬁtting at the bottom of a valve, the removal of which permits draining and ﬂushing the

body cavity.

DREDGE PUMP : A dredge pump is a submersible, centrifugal pump capable of handling high solids

concentrations and is typically used for clearing out and/or deepening harbors and waterways. The mate-

rial being moved (i.e.) sand, dirt, soil, etc.) is carried away along with the water it is suspended in.

DRIVE PINS : The two pins which ﬁt into the bottom of a ball valve stem and engage corresponding

holes in the ball. As the operator turns the stem, the drive pins turn the ball.

DROOP : A drop in set (outlet) pressure of a regulator or control valve due to the travel of its valve or pop-

496

Chapter 25. Glossary

pet, as the required ﬂow increases from low to maximum. A slight change in the control spring length due

to the valve travel, will result in spring force variations, translating into a change of set (outlet) pressure.

DROP MANHOLE : A main line or house service line lateral entering a manhole at a higher elevation

than the main ﬂow line or channel. If the higher elevation ﬂow is routed to the main manhole channel

outside of the manhole, it is called an “outside drop.” If the ﬂow is routed down through the manhole

barrel, the pipe down to the manhole channel is called an “inside drop.”

DRY WELL : A dry room or compartment in a lift station, near or below the water level, where the pumps

are located.

DUCKWEED : A water plant with single small leaf that ﬂoats and accumulates on the surface of lagoons.

E. COLI OR ESCHERICHIA COLIFORM : A species of bacteria found in large numbers in the intestinal

tract of warm–blooded animals.

EASEMENT : Legal right to use the property of others for a speciﬁc purpose. For example, a utility

company may have a ﬁve–foot easement along the property line of a home. This gives the utility the legal

right to install and maintain a sewer line within the easement.

EFFICIENCY : A ratio of total power output to the total power input, expressed as a percent.

EFFLUENT : Wastewater or other liquid—raw (untreated), partially, or completely treated—ﬂowing

FROM a reservoir, basin, treatment process, or treatment plant.

ELASTOMER : A natural or synthetic elastic material, often used for o–ring seals. Typical materials are

viton, buna–n, EPDM (ethylene propylene dimonomer), etc.

ELECTRO DEIONIZATION : Water treatment technology that utilizes an electricity, ion exchange mem-

branes and resin to deionize water and separate dissolved ions (impurities) from water.

ELEVATION : The height to which something is elevated, such as the height above sea level.

ELUTRIATION : The washing of digested sludge with fresh water, plant efﬂuent or other wastewater. The

goal is to remove ﬁne particles and/or the alkalinity in the sludge. This process reduces the demand for

conditioning chemicals and improves settling or ﬁltering characteristics of the sludge.

ELUTRIATION : The washing of digested sludge with fresh water, plant efﬂuent or other wastewater. The

goal is to remove ﬁne particles and/or the alkalinity in the sludge. This process reduces the demand for

conditioning chemicals and improves settling or ﬁltering characteristics of the sludge.

EMERGENCY SEAT SEAL : To obtain tight shut off in an emergency situation, a sealant can be injected

into a specially designed groove in the seat rings. Available for most ball valves and gate valves.

EMULSION BREAKER (DEMULSIFIER) : Chemical additive that destroys the emulsifying characteris-

tics water. Typically used separate stabilized (emulsiﬁed) oil in water.

ENTHALPY : A thermodynamic property of a ﬂuid. The enthalpy of a ﬂuid consist of the energy associ-

ated with the ﬂuid at a microscopic level (related to the temperature of the ﬂuid) plus the energy present in

the form of pressure at the inlet and outlet of a system.

EODD : EODD stands for Electrically Operated Double Diaphragm (pumps). These are diaphragm pumps

driven either directly or indirectly with an electric motor. Offering many of the same advantages as

AOD/AODD pumps with less noise and no compressed air/gas requirements. They cannot run against

a closed discharge, which air operated models can, except Graco’s e–Series models.

EPA : United States Environmental Protection Agency.

497

EQUALIZING BASIN : A holding basin in which variations in ﬂow and composition of a liquid are

averaged. Such basins are used to provide a ﬂow of reasonably uniform volume and composition to a

treatment unit. Also called a balancing reservoir.

EQUALIZING BASIN : A holding basin in which variations in ﬂow and composition of a liquid are

averaged. Such basins are used to provide a ﬂow of reasonably uniform volume and composition to a

treatment unit. Also called a balancing reservoir.

ESCHERICHIA COLIFORM : A species of bacteria found in large numbers in the intestinal tract of

warm–blooded animals.

ESDV : EMERGENCY SHUT DOWN VALVES – A valve or a system of valves which, when activated,

initiate a shut–down of the plant, process, or platform they are tied to.

ESTUARY : Bodies of water that are located at the lower end of a river and are subject to tidal ﬂuctua-

tions.

ESTUARY : Bodies of water that are located at the lower end of a river and are subject to tidal ﬂuctua-

tions.

EUTROPHICATION : The increase of nutrient levels of a lake or other body of water; this usually causes

in increase in the growth of aquatic animal and plant life.

EUTROPHICATION : The increase of nutrient levels of a lake or other body of water; this usually causes

in increase in the growth of aquatic animal and plant life.

EVAPOTRANSPIRATION : 1) The process by which water vapor passes into the atmosphere from living

plants. Also called Transpiration. 2) The total water removed from an area by transpiration (living plants)

and by evaporation from soil, snow and water surfaces.

EVAPOTRANSPIRATION : 1) The process by which water vapor passes into the atmosphere from living

plants. Also called Transpiration. 2) The total water removed from an area by transpiration (living plants)

and by evaporation from soil, snow and water surfaces.

EXFILTRATION : Liquid wastes and liquid–carried wastes which unintentionally leak out of a sewer pipe

system and into the environment.

EXPANDING GATE VALVE : A gate valve that is comprised of a separate gate and segment that as the

valve operates the gate and segment move without touching the seats, permitting the valve to be opened

and closed without wear. In the closed position the gate and segment are forced against the seat. Con-

tinued downward movement of the gate causes the gate and segment to expand against the seats. When

the valve reaches its full open position, the gate and segment seal off against the seats while the ﬂow is

isolated from the valve body.

EXTENDED AERATION : A modiﬁcation of the activated sludge process which provides for aerobic

sludge digestion within the aeration system.

EXTENDED BDV (BLOW DOWN VALVE) : Used on buried valves where the drain plug is inaccessi-

ble. Instead, a line is piped above grade, terminating in a small valve. Line pressure is used to blowout

condensates and other material which settles out in the bottom of the body cavity.

EXTENSIONS : The equipment applied to buried valves to provide above grade accessibility to operating

gear, blowdown and seat lubrication systems.

FACE: TO–FACE – The overall dimension from the inlet face of a valve to the outlet face of the valve

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Chapter 25. Glossary

(one end to the other). This dimension is governed by ASME B16.10 and API–6D to ensure that such

valves are mutually inter changeable, regardless of the manufacturer.

FACULTATIVE : A combination of both aerobic and anaerobic conditions. Facultative cells have both aer-

obic and anaerobic zones. Facultative bacteria are able to exist in both aerobic and anaerobic conditions.

A facultative pond is commonly used to treat wastewater ﬂows in small communities, It has an upper

aerobic zone, a lower anaerobic zone, and algae provide most of the oxygen for the bacteria.

FACULTATIVE ORGANISMS : Organisms that can survive and function in the presence or absence of

free, elemental oxygen. Basically, organisms that can switch from aerobic or anaerobic depending on its

environment. 227

FACULTATIVE POND (ALSO KNOWN AS A WASTEWATER TREATMENT POND OR LAGOON)

: The most common type of treatment pond used for treating domestic wastewater. The upper portion is

aerobic, while the bottom layer is anaerobic. Algae supply most of the oxygen in the aerobic layer.

FAIL SAFE VALVE : A valve designed to fail in a preferred position (open or closed) in order to avoid an

undesirable consequence in a piping system.

FAIR LEAD PULLEY : A pulley that is placed in a manhole to guide TV camera electric cables and the

pull cable into the sewer when inspecting pipelines.

FATS OIL & GREASE : a term to describe wastewater contaminants that are commonly found in food or

petroleum based efﬂuents. EPA test 1664 is typically used to measure FOG.

FATS, OILS, & GREASE : Food industry byproducts that can cause signiﬁcant problems for sewer sys-

tems. This pollutant includes both animal/vegetable and petroleum sources and can be regulated sepa-

rately by these fractions. It is important to know which fraction is regulated and what analytical method is

being used to get accurate results.

FECAL COLIFORM : A type of bacteria found in the bodily discharges of warm–blooded animals. Used

as an indicator organism.

FERMENTATION : A process of decomposition of organic solid materials by bacteria and other biologi-

cal actions.

FERRIC CHLORIDE : Metal salt (FeCl3) commonly used as an coagulant in water clariﬁcation and as

etching agent in chemical–etching.

FIELD SERVICEABLE : A statement indicating that normal repair of the valve or replacement of operat-

ing parts can be accomplished in the ﬁeld without return to the manufacturer.

FILAMENTOUS : The property of growing in long strings, or ﬁlaments. Algae and bacteria have ﬁla-

mentous forms. Algae ﬁlaments can clog up equipment and be a nuisance in receiving waters. Bacterial

ﬁlaments area common cause of bulking in activated sludge.

FILAMENTOUS ORGANISMS : Organisms that grow in a thread or ﬁlamentous form. Common types

are Thiothrix and Actinomycetes. A common cause of sludge bulking in the activated sludge process.

FILTER PRESS : Solids dewatering device that uses pressure differential applied to sludge within a series

of plates with ﬁlter clothes. The plates (with clothes in them) are arranged in a plate pack with the sludge

ﬁlling chambers created by recesses within each plate. Filter presses are often called Plate and Frame or

Recessed Chamber Filter Presses.

FIRE GATE : A gate or ball valve which is positioned in a pipeline at the entrance to a compressor station.

499

This valve is closed in case of ﬁre in the compressor station. Closing the valve prevents the gas in the

pipeline from feeding the ﬁre.

FIRE SAFE : A statement associated with a valve design which is capable of passing certain speciﬁed

leakage and operational tests after exposure to ﬁre. Must be referenced to a particular speciﬁcation.

FLEXIBLE TUBE VALVE : A special valve using a ﬂexible sleeve or tube which acts as the closure

element. Pressure applied to the jacket space surrounding the outside of the tube, controls the opening and

closing of the valve.

FLOAT LINE : A length of rope or heavy twine attached to a ﬂoat, plastic jug or parachute to be carried

by the ﬂow in a sewer from one manhole to the next. This is called “stringing the line” and is used for

pulling through winch cables, such as for a bucket machine work or closed–circuit television work.

FLOAT VALVE : A valve which automatically opens or closes as the level of a liquid changes. The valve

is operated mechanically by a ﬂoat which rests on the top of the liquid.

FLOATING BALL : A ball valve having a non–trunnion mounted ball. The ball is free to ﬂoat between

the seat rings, and thus causes higher torques.

FLOC : The agglomeration of smaller particles in a gelatinous mass that can be more easily removed

from the liquid than the individual small particles. Clumps of bacteria and particles or coagulants and

impurities that have come together and formed a cluster. Found in aeration tanks, secondary clariﬁers and

chemical precipitation processes.

FLOCCULANT : high molecular weight polyelectrolyte water soluble organic polymer designed to

agglomerate solids in water substrates. Characteristics of ﬂocculants in water treatment are determined by

their molecular weight, charge type (anionic, nonionic or cationic) and charge density.

FLOCCULATION : the agglomeration of settleable solids through a bridging mechanism to produce

larger particle that is more easily separated from water.

FLOODED SUCTION : In a ﬂooded suction system, the liquid ﬂows to the pump inlet from an elevated

source by means of gravity. This is generally recommended for centrifugal pumps.

FLOTATION : (1) The stress or forces on a pipeline or manhole structure below a water table which tend

to lift or ﬂoat the pipeline or manhole structure. (2) The process of raising suspended matter to the surface

of the liquid in a tank where it forms a scum layer that can be removed by skimming. The suspended

matter is raised by aeration, the evolution of gas, the use of chemicals, electrolysis, heat or bacterial

decomposition.

FLOW : A measure of the liquid volume capacity of a pump. Given in gallons per minute/hour (gpm,

gph), liters per minute/hour (L/min, L/hour), milliliters per minute (mL/min), cubic meters per hour

(m3/h) and other rarely used measurements.

FLOW : The continuous movement of a liquid from one place to another.

FLOW ISOLATION : A procedure used to measure inﬂow and inﬁltration (I/I). A section of sewer is

blocked off or isolated and the ﬂow from the section is measured.

FLUME : (1) An open conduit of wood, masonry, metal, or plastic constructed on a grade and sometimes

elevated. (2) A ﬂow rate measurement device.

FLUSHER BRANCH : A line built speciﬁcally to allow the introduction of large quantities of water to

the collection system so the lines can be “ﬂushed out” with water. Also installed to provide access for

500

Chapter 25. Glossary

equipment to clear stoppages in a sewer.

FLUSHING : The removal of deposits of material which have lodged in sewers because of inadequate ve-

locity of ﬂows. Water is discharged into the sewers at such rates that the larger ﬂow and higher velocities

are sufﬁcient to remove the material.

FLUX : the permeate rate per unit area of membrane surface. Typical units are gals per foot of membrane

per day (gfd) or liters of permeate per square meter of membrane per hour (lmh).

FOOD: TO–MASS RATIO (F/M) – An activated sludge process–control calculation based upon the

amount of food (BOD5 or COD) available per pound of mixed liquor volatile suspended solids.

FORCE MAIN : A pipe that carries wastewater under pressure from the discharge side of a pump to a

point of gravity ﬂow downstream.

FREE AVAILABLE RESIDUAL CHLORINE : That portion of the total available chlorine residual com-

posed of dissolved chlorine gas (Cl2), hypochlorous acid (HOCl), and/or hypochlorite ion (OCl–) remain-

ing in water after chlorination.

FREE OXYGEN : Oxygen can be dissolved in water as the soluble gas O2 when it is called free oxygen

and measured as dissolved oxygen.

FREEBOARD : The vertical distance from the normal water surface to the top of the conﬁning wall.

FRICTION : The force produced as a reaction to movement. All ﬂuids produce friction when they are

in motion. The higher the ﬂuid viscosity, the higher the friction force for the same ﬂow rate. Friction

is produced internally as one layer of ﬂuid moves with respect to another and also at the ﬂuid/surface

interface.

FRICTION HEAD : The pressure expressed in pounds per square inch or feet of liquid needed to over-

come the resistance to the ﬂow in pipes and ﬁttings.

FRICTION HEAD DIFFERENCE : The difference in head required to move a mass of ﬂuid from one

position to another at a certain ﬂow rate.

FRICTION LOSS : The head lost by water ﬂowing in a stream or conduit as the result of the disturbances

set up by the contact between the moving water and its containing conduit and by intermolecular friction.

FULL BORE (FULL PORT) : Describes a valve in which the bore (port) is nominally equal to the bore of

the connecting pipe.

FULL OPENING : Describes a valve whose bore (port) is nominally equal to the bore of the connecting

pipe.

GAC : Granular Activated Carbon – A material used to absorb organics from wastewater. This char-

coal–like material can be used in ﬁltration systems to remove solvent contamination.

GATE : The closure element of a gate valve (sometimes called wedge or disc).

GATE VALVE : A straight–thru pattern valve whose closure element is a wedge or parallel–sided slab, sit-

uated between two ﬁxed seating surfaces, with means to move it in or out of the ﬂow stream in a direction

perpendicular to the pipeline axis.

GLAND FOLLOWER OR GLAND FLANGE : The component used to hold down or retain the gland in

the stufﬁng box.

GLAND OR GLAND BUSHING : That part of a valve which retains or compresses the stem packing in a

stufﬁng box (where used) or retains a stem O–ring, lip seal, or stem O–ring bushing. Sometimes manually

501

adjustable.

GLAND PLATE : The plate in a valve which retains the gland, gland bushing or stem seals and some-

times guides the stem.

GLOBE VALVE : A valve whose closure element is a ﬂat disc or conical plug sealing on a seat which is

usually parallel to the ﬂow axis. Can be used for throttling services.

GO : GEAR OPERATED – The actuation of a valve thru a – ear set which multiplies the torque applied to

the valve stem.

GRAB SAMPLE : A single sample of water collected at a particular time and place which represents the

composition of water only at that time and place.

GRADE : (1) The elevation of the invert (or bottom) of a pipeline, canal, culvert, sewer, or similar conduit.

(2) The inclination of slope of a pipeline, conduit, stream channel, or natural ground surface; usually ex-

pressed in terms of the ratio or percentage of number of units of vertical rise or fall per unit of horizontal

distance. A 0.5 percent grade would be a drop of one–half foot per hundred feet of pipe.

GRAVITY FLOW : Water or wastewater ﬂowing from a higher elevation to a lower elevation due to

the force of gravity. The water does not ﬂow due to energy provided by a pump. Wherever possible,

wastewater collection systems are designed to use the force of gravity to convey waste liquids and solids.

GREASE : In wastewater, a group of substances, including fats, waxes, free fatty acids, calcium and

magnesium soaps, mineral oils, and certain other non–fatty materials.

GREASE : In a collection system, grease is considered to be the residues of fats, detergents, waxes, free

fatty acids,

GREASE BUILDUP : Any point in a collection system where coagulated and solidiﬁed greases accu-

mulate and build up. Many varieties of grease have high adhesive characteristics and collect other solids,

forming restrictions and stoppages in collection systems.

GREASE FITTING : A ﬁtting through which lubricant or sealant is injected.

GREASE TRAP : A receptacle designed to collect and retain grease and fatty substances usually found in

kitchens or from similar wastes. It is installed in the drainage system between the kitchen or other point

of production of the waste and the building wastewater collection line. Commonly used to control grease

from restaurants.

GREEN ALGAE : The common forms of algae in an aerobic lagoon environment. Green algae are essen-

tial for lagoon treatment.

GRINDER PUMP : A grinder pump is a waste management device. Waste from water–using household

appliances (toilets, bathtubs, washing machines, etc.) ﬂows through the home’s pipes into the grinder

pump’s holding tank. Once the waste inside the tank reaches a certain level, the pump will turn on, grind

the waste into ﬁne slurry, and pump it to the central sewer system.

GRIT : The heavy mineral material present in wastewater such as sand, coffee grounds, eggshells, gravel

and cinders. Grit tends to settle out at ﬂow velocities below 2 ft. /sec,

GRIT CATCHER : A chamber usually placed at the upper end of a depressed collection line or at other

points on combined or storm water collection lines where wear from grit is possible. The chamber is sized

and shaped to reduce the velocity of ﬂow through it and thus permit the settling out of grit.

GRIT REMOVAL : Grit removal is accomplished by providing an enlarged channel or chamber which

502

Chapter 25. Glossary

causes the ﬂow velocity to be reduced and allows the heavier grit to settle to the bottom of the channel

where it can be removed.

GRIT REMOVAL : Grit removal is accomplished by providing an enlarged channel or chamber which

causes the ﬂow velocity to be reduced and allows the heavier grit to settle to the bottom of the channel

where it can be removed.

GRIT TRAP : A permanent structure built into a manhole (or other convenient location in a collection

system) for the accumulation and easy removal of grit.

HAND WHEEL : A wheel–shaped valve operating device intended to be grasped with one or both hands

which allows turning the valve stem or operator shaft to which it is attached.

HARD FACING : A surface preparation in which an alloy is deposited on a metal surface usually by weld

overlay to increase resistance to abrasion and or corrosion.

HARD WATER : Water having a high concentration of calcium and magnesium ions.

HARDNESS : It is typically the concentration of calcium and magnesium salts in water. However, it may

include other metal salts such as Al, Mn, Sr and Zn. Normally measured as CaCO3 equivalents.

HEAD : Refers to the pressure produced by a vertical column of ﬂuid. It is a measure of pressure, ex-

pressed in feet of head for pumps. Water is used as the default where 10 meters (33.9 ft.) of water equals

one atmosphere (14.7 psi. or 1 bar).

HEAD : The vertical distance (in feet) equal to the pressure (in psi) at a speciﬁc point. The pressure head

is equal to the pressure in psi times 2.31 ft/psi.

HEADWORKS : The part of wastewater treatment where the raw sewage is ﬁrst received for treatment.

Preliminary treatment occurs at the Headworks.

HETEROTROPH : a type of bacteria that uses organic matter for energy and can use free oxygen, nitrates

or sulfates as and oxygen source for respiration.

HUBS : The end connection tubes on a gate valve.

HUMAN MACHINE INTERFACE (HMI) : An interface such as a display screen that permits interaction

between a human being and a device/equipment or process. HMI allows for the visual display of status

and condition of a particular process or equipment.

HWO : HANDWHEEL OPERATED – A valve on which the handwheel drives the stem directly to oper-

ate the valve.

HYDRAULIC LOADING : The ﬂow of water per acre of surface area.

HYDRAULIC MOTOR ACTUATOR (OPERATOR) : A device by which rotation of a hydraulically

powered motor is converted into mechanical motion.

HYDROGEN SULFIDE : A very odorous and poisonous gas. Commonly known as rotten egg gas. It is a

combined form of hydrogen and sulfur with the formula H2S.

HYDROGEN SULFIDE GAS (H2S) : Hydrogen sulﬁde is a gas with a rotten egg odor. This gas is pro-

duced under anaerobic conditions. Hydrogen sulﬁde is particularly dangerous because it dulls your sense

of smell so that you do not notice it after you have been around it for a while and because the odor is

not noticeable in high concentrations. The gas is very poisonous to your respiratory system, explosive,

ﬂammable and colorless.

HYDROLOGIC CYCLE : The process of evaporation of water into the air and its return to earth by

503

precipitation. (Also called the WATER CYCLE)

HYPOCHLORINATORS : Chlorine pumps, chemical feed pumps or devices used to dispense chlorine

solutions made from hypochlorites into the water being treated.

IMPELLER : The rotating element of a centrifugal pump which imparts movement and pressure to a ﬂuid.

INCLINED PLATE CLARIFIER : a solids / liquid separation device (clariﬁer) that is ﬁlled with parallel

(sometimes called Lamella) plates that are inclined at an angle between 45 and 55 degrees. The plates

reduce (compared to a gravity clariﬁer) the foot–print required to properly settle solids.

INCREMENTAL SEAT TEST : The leakage testing of valve seats in an assembled valve by increasing the

applied pressure in prescribed pressure steps.

INDUSTRIAL WASTEWATER : Wastes associated with industrial manufacturing processes. 229

INFECTION : Introduction of presence of pathogenic organisms in potable water supply.

INFECTION : Introduction of the presence of pathogenic organisms in potable water supply. This is

determined in two ways:

INFILTRATION : The gradual ﬂow of water into the soil; also, the ﬂow of groundwater as seepage into a

sewer system.

INFILTRATION HEAD : The distance from a point of inﬁltration leaking into a collection system to

the water table elevation. This is the pressure of the water being forced through the leak in the collection

system.

INFILTRATION/INFLOW : The total quantity of water from both inﬁltration and inﬂow without distin-

guishing the source. Abbreviated I&I or I/I.

INFLATABLE PIPE STOPPER : An inﬂatable ball or bag used to form a plug to stop ﬂows in a sewer

pipe.

INFLOW : Water discharged into a sewer system and service connections from sources other than regular

connections. This includes ﬂow from yard drains, foundation drains and around manhole covers. Inﬂow

differs from inﬁltration in that it is a direct discharge into the sewer rather than a leak in the sewer itself.

INFLUENT : Wastewater or other liquid—raw (untreated) or partially treated—ﬂowing into a reservoir,

basin, treatment process, or treatment plant.

INLET : (1) A surface connection to a drain pipe. (2) A chamber for collecting storm water with no well

below the outlet pipe for collecting grit. Often connected to a CATCH BASIN or a “basin manhole” with

a grit chamber.

INLET PORT : That end of a valve which is connected to the upstream pressure zone of a ﬂuid system.

INNER SEAT RING : The inner part of a two–piece valve seat assembly.

INORGANIC : Material such as salts, metals, and all other substances of mineral origin.

INORGANIC : Material such as sand, salt, iron, calcium salts and other mineral materials and other than

of plant or animal origin or of carbon compounds (ORGANIC).

INORGANIC MATERIAL : Material that will not respond to biological action (sand, cinders, stone).

Non–volatile fraction of solids.

INORGANIC MATERIAL : Material that will not respond to biological action (sand, cinders, stone).

Non–volatile fraction of solids.

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Chapter 25. Glossary

INSERTION PULLER : A device used to pull long segments of ﬂexible pipe material into a sewer line

when sliplining to rehabilitate a deteriorated sewer.

INSIDE: OUT AIR SEAT TEST – A pressure test that can be performed only on independent seating

trunnion mounted ball valves. By closing the valve and pressurizing the body cavity, all of the seals in an

independent seating ball valve can then be pressure tested.

INSITUFORM : A method of installing a new pipe within an old pipe without excavation. The process

involves the use of a polyester–ﬁber felt tube, lined on one side with polyurethane and fully impregnated

with a liquid thermal setting resin.

INSPECTION TELEVISION EQUIPMENT : Television equipment that is superior to standard commer-

cial quality, providing 600 to 650 lines of resolution, and designed for industrial inspection applications.

INTEGRATED FIXED FILM ACTIVATED SLUDGE (IFAS) : A suspended growth system that pro-

vides additional biomass within a biological wastewater treatment system to meet more stringent efﬂuent

parameters or increased loadings.

INTERNAL PRESSURE RELIEF : A self relieving feature in non–independent seating valves that auto-

matically relieves excessive internal body pressure caused by sudden changes in line pressures. By means

of the piston effect principal the excessive body pressure will move the seat away from its seating surface

and relieve it to the lower pressure side.

INVERT : The lowest point of the channel inside a pipe or manhole.

INVERTED SIPHON : A pressure pipeline used to carry wastewater ﬂowing in a gravity collection sys-

tem under a depression such as a valley or roadway or under a structure such as a building. 230

ION EXCHANGE : Process that removes dissolved ions from solution of a certain charge by absorption

onto a resin that releases (exchanges) an ion of the same charge.

ISRS : Inside screw, rising stem – common term for any valve design in which the stem threads are ex-

posed to the ﬂuid below the packing and the stem rises up through the packing when the valve is opened.

KEY MANHOLE : In collection system evaluation, a key manhole is one from which reliable or speciﬁc

data can be obtained.

KEY STOP : A method of restricting the travel of a ball valve from fully open to fully closed. The stem

key bears against the ends of an arc machined in the adaptor plate.

KITE : A device for hydraulically cleaning sewer lines. Resembling an airport wind sock and constructed

of canvas–type material, the kite increases the velocity of a ﬂow at its outlet to wash debris ahead of it.

LAMINAR : A distinct ﬂow regime that occurs at low Reynolds number (Re < 2000). It is characterized

by particles in successive layers moving past one another in a well behaved manner (little to no turbu-

lence).

LAMPING : Using reﬂected sunlight or a powerful light beam to inspect a sewer between two adjacent

manholes. The light is directed down the pipe from one manhole. If it can be seen from the next manhole,

it indicates that the line is open and straight.

LATERAL : (See LATERAL SEWER)

LATERAL CLEANOUT : A capped opening in a building lateral, usually located on the property line,

through which the pipelines can be cleaned.

LATERAL SEWER : A sewer that discharges into a branch or other sewer and has no other common

505

sewer tributary to it. Sometimes called a “street sewer” because it collects wastewater from individual

homes.

LEAD : A heavy metal commonly regulated by wastewater discharge permits. Other heavy metals include

– Arsenic (As), Cadmium (Cd), Chromium(Cr), Copper (Cu), Nickel (Ni), and Zinc (Zn).

LEVER : A handle type operating device for quarter–turn valves.

LIFT STATION : A wastewater pumping station that lifts the wastewater to a higher elevation when

continuing the sewer at reasonable slopes would involve excessive depths of trench. Also, an installation

of pumps that raise wastewater from areas too low to drain into available sewers. These stations may be

equipped with air–operated ejectors or centrifugal pumps. Sometimes called a PUMP STATION, but this

term is usually reserved for a similar type of facility that is discharging into a long FORCE MAIN, while a

lift station has a discharge line or force main only up to the downstream gravity sewer.

LIFTING LUGS : Lugs provided on large ball, gate, and check valves, for lifting and positioning valves.

Also called lifting eyes.

LIMIT SWITCH : An electrical device providing a signal to a remote observation station indicating when

the valve is in the fully open or fully closed position. Usually a component of a valve operator.

LIQUID END : A term referring to the side (or parts) of the pump that come into contact with the process

ﬂuid. This applies to any air operated pump including Air Operated Diaphragm Pumps, air operated piston

pumps and air operated drum pumps.

MAGNETIC DRIVE : Also referred to as a Mag–drive. This is a method of connecting the motive force

to the pump which uses a series of magnets coupled together, with a containment chamber separating

them. Magnetic Drives keep the ﬂuid sealed from atmosphere and other environmental factors and elim-

inate the need for seals and seal maintenance. Special considerations must be taken into account when

specifying a Mag–Drive pump or mixer. Ask Springer Pumps for more information.

MAIN LINE : Branch or lateral sewers that collect wastewater from building sewers and service lines.

MAIN SEWER : A sewer line that receives wastewater from many tributary branches and sewer lines and

serves as an outlet for a large territory or is used to feed an intercepting sewer.

MANDREL : (1) A special tool used to push bearings in or to pull sleeves out. (2) A gage used to measure

for excessive deﬂection in a ﬂexible conduit.

MANHOLE : An opening in a sewer provided for the purpose of permitting operators or equipment to

enter or leave a sewer.

MANHOLE ELEVATION : The height (elevation) of the invert or lowest point in the bottom of a manhole

above mean sea level.

MANHOLE FLOW : (1) The depth or amount of wastewater ﬂow in a manhole as observed at any se-

lected time. (2) The total or the average ﬂow through a manhole in gallons on any selected time interval.

231

MANHOLE INFILTRATION : Groundwater that seeps or leaks into a manhole structure.

MANHOLE INFLOW : Surface waters ﬂowing into a manhole, usually through the vent holes in the

manhole lid.

MANHOLE INVERT : The lowest point in a trough or ﬂow channel in the bottom of a manhole.

MANHOLE LID : The heavy cast–iron or forged–steel cover of a manhole. The lid may or may not have

506

Chapter 25. Glossary

vent holes.

MANHOLE LID DUST PAN : A sheet metal or cast–iron pan located under a manhole lid. This pan

serves to catch and hold pebbles and other debris falling through vent holes, preventing them from getting

into the pipe system.

MANHOLE VENTS : One or a series of one–inch diameter holes through a manhole lid for purposes of

venting dangerous gases found in sewers.

MASKING AGENTS : Substances used to cover up or disguise unpleasant odors. Liquid masking agents

are dripped into the wastewater, sprayed into the air, or evaporated (using heat) with the unpleasant fumes

or odors and then discharged into the air by blowers to make an undesirable odor less noticeable.

MASKING AGENTS : Substances used to cover up or disguise unpleasant odors. Liquid masking agents

are dripped into the wastewater, sprayed into the air, or evaporated (using heat) with the unpleasant fumes

or odors and then discharged into the air by blowers to make an undesirable odor less noticeable.

MBR : Membrane Bio Reactor – A wastewater treatment technology that combines biological treatment

with physical treatment involving membrane ﬁltration. It’s used primarily to treat BOD, COD, and sus-

pended solids to very low levels where efﬂuent may be able to be reused or recycled.

MEAN CELL RESIDENCE TIME (MCRT) : The average length of time a mixed liquor suspended solids

particle remains in the activated sludge process. May also be known as sludge retention time.

MECHANICAL AERATION : Method of aeration.

MECHANICAL AERATION : The use of machinery to mix air and water so that oxygen can be absorbed

into the water.

MECHANICAL CLEANING : Clearing pipe by using equipment that scrapes, cuts, pulls or pushes

the material out of the pipe. Mechanical cleaning devices or machines include bucket machines, power

rodders and hand rods.

MECHANICAL PLUG : A pipe plug used in sewer systems that is mechanically expanded to create a

seal.

MECHANICAL SEAL : In a valve, a shut off that is accomplished by a mechanical means rather than

with ﬂuid or line pressure. The wedging action of a gate against the seats or the seat springs pushing the

seat against the ball or gate are examples of mechanical seals in a valve.

MEMBRANE : Material layer that is a selective barrier between two phases and remains impermeable to

speciﬁc particles, molecules or substances.

MEMBRANE BIO REACTOR : Aerobic biological wastewater treatment process that utilizes membrane

ﬁltration (rather than clariﬁcation) for solids / liquid separation. The membrane ﬁlters (ultra ﬁlters) can be

either submerged or external to the biological mixed liquor tanks.

MEMBRANE BIOREACTOR (MBR) : Biological wastewater treatment process where a selected mem-

brane is integrated with a biological process to act as a suspended growth bioreactor.

MERCURY (HG) : A metal which remains liquid at room temperature. This property makes it useful

when used in a thin vertical glass tube since small changes in pressure can be measured as changes in the

mercury column height. The inch of mercury is often used as a unit for negative pressure.

METAL: TO–METAL SEAL – The seal produced by metal–to–metal contact between the sealing face of

the seat ring and the closure element, without beneﬁt of a synthetic seal.

507

METERING PUMP : Pumps used for precise introductions of chemicals into a tank, existing ﬂuid stream

or some other liquid handling equipment. Types of pumps for these include Diaphragm Pumps (AOD

or EOD), Peristaltic Pumps, Hose Pumps, Gear Pumps, Bellows Pumps, Piston Pumps and other less

commonly used pump types.

METHANE : A combustible gas produced during anaerobic fermentation of organic matter, such as by

anaerobic digestion of wastewater solids.

MGO : MANUAL GEAR OPERATOR – A gear operator that is operated manually (with a handwheel).

MGWPCS PERMIT : Montana Groundwater Pollution Control System permit. This permit is issued to

owners/operators of potential sources of pollution to state ground waters.

MICRO FILTRATION : Type of membrane ﬁltration that separates suspended solids and solutes of high

molecular from a liquid and low molecular weight solutes. This separation process is used in industry

processes, water treatment and research for purifying and concentrating macromolecular solutions.

MICRO: Organisms – Microscopic plans and animals such as bacteria, molds, protozoa, algae, and small

metazoa.

MICRO: Organisms – Microscopic plants and animals such as bacteria, molds, protozoa, algae, and small

metazoa.

MICROORGANISMS : Microscopic living organisms.

MICROORGANISMS : Very small organisms that can be seen only through a microscope. Some microor-

ganisms use the wastes in wastewater for food and thus remove or alter much of the undesired matter.

MILLIGRAMS PER LITER, MG/L : A measure of the concentration by weight of a substance per unit

volume. One thousandth of a gram in one liter. One mg/L is equal to one part per million (ppm).

MIXED BED ION EXCHANGE : Anion exchange process that uses a mixture of cation and anion resin

combined in a single ion exchange column. With proper pretreatment, product water puriﬁed from a

single pass through a mixed bed ion exchange column is the purest that can be made.

MIXED LIQUOR : When the activated sludge in an aeration tank is mixed with primary efﬂuent or the

raw wastewater and return sludge, this mixture is then referred to as mixed liquor as long as it is in the

aeration tank. Mixed liquor may also refer to the contents of mixed aerobic or anaerobic digesters.

MIXED LIQUOR SUSPENDED SOLIDS (MLSS) : The milligrams of suspended solids per liter of

mixed liquor that are combustible at 550 degrees Centigrade. An estimate of the quantity of MLSS to be

wasted from the aeration tank of an extended aeration plant may be determined by the rate of settling and

centrifuge tests on the sludge solids.

MIXED LIQUOR VOLATILE SUSPENDED SOLIDS (MLSS) : The concentration of organic matter in

the mixed liquor suspended solids.

MIXED LIQUOR VOLATILE SUSPENDED SOLIDS (MLVSS) : The organic or volatile suspended

solids in the mixed liquor of an aeration tank. This volatile portion is used as a measure or indication of

the microorganisms present.

MIXED MEDIA GRAVITY FILTER : A ﬁlter using more than one ﬁltering media (such as coal and

sand).

MIXER, SUBMERSIBLE : A submersible mixer is a mechanical device that is used to mix sludge tanks

and other liquid volumes. Submersible mixers are often used in sewage treatment plants to keep solids in

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Chapter 25. Glossary

suspension in the various process tanks and/or sludge holding tanks. The submersible mixer is operated

by an electric motor, which is coupled to the mixer’s propeller, either direct–coupled or via a planetary

gear–reducer. The propeller rotates and creates liquid ﬂow in the tank, which in turn keeps the solids in

suspension. The submersible mixer is typically installed on a guide rail system, which enables the mixer

to be retrieved for periodic inspection and preventive maintenance.

MIXER, VERTICAL : This style of mixer uses an extended shaft between the motor and mixing blade

such that the motor is above and out of the liquid. These also incorporate a gear reducer to slow the

speed of the mixing blades to achieve the desired mixing/tank turnover rate. Vertical mixers are often

used in reactor vessels to ensure thorough chemical reactions or to mix different ingredients together in

food/beverage applications.

MOISTURE CONTNET : The amount of water per unit weight of bio solids.

MONITORING WELL : Wells used to collect groundwater samples for analysis to determine the amount,

type, and spread of contaminants in groundwater. Speciﬁc design is often determined by the Water Protec-

tion Bureau at MT DEQ.

MOVING BED BIO REACTOR : Aerobic biological wastewater treatment process that utilizes the ﬁxed

ﬁlm (media) process. High surface area media is suspended in biological mixed liquor and bacteria grow

on the media (attached growth) surface. A clariﬁer is typically used downstream for solid / liquid separa-

tion.

MOVING BED BIOREACTOR (MBBR) : Method to biologically treat wastewater by circulating moving

media in an aerobic sludge environment.

MPDES PERMIT : Montana Pollutant Discharge Elimination System permit. This permit lists the condi-

tions that must be met before treatment plants can discharge an efﬂuent into state

MULTI: Stage Pump – The multi–stage pump is used for clean, clear liquids requiring signiﬁcant dis-

charge pressure. A multi–stage pump is nothing more than a standard centrifugal pump with the discharge

of the initial volute discharging directly into the suction of the next volute. The numerous “volutes” are

all internal to the pump and many times the volutes are hard to spot individually. The number of stages is

dependent on the desired Total Discharge Head required by the application. These types of pump can be

either horizontal or vertical conﬁguration. Commonly used for boiler feed.

NEEDLE VALVE : A type of small valve, used for ﬂow metering, having a tapered needle–point plug or

closure element and a seat having a small oriﬁce.

NEGATIVE PRESSURE : Pressure that is less than the pressure in the external environment.

NEPHELOMETER TURBIDITY UNITS : It is a measurement of the clarity (turbidity) of a liquid.

NEPHELOMETRIC TURBIDITY UNIT : The standard unit to measure turbidity or how cloudy water is.

It’s used as a visual indicator for how well a wastewater treatment system is working.

NET POSITIVE SUCTION HEAD (NPSH.) : The head in feet of water absolute as measured or cal-

culated at the pump suction ﬂange, less the vapor pressure (converted to feet of water absolute) of the

ﬂuid.

NET POSITIVE SUCTION HEAD AVAILABLE (NPSHA) : The NPSHa available to prevent cavita-

tion of the pump. To calculate the NPSHa, take the (Static Suction Head) plus (Suction Vessel Surface

Pressure Head) minus (vapor pressure of ﬂuid) minus (friction losses in the suction).

NET POSITIVE SUCTION HEAD REQUIRED (NPSHR) : The NPSHr to stop a pump from cavitating.

509

The NPSHr is generally supplied to you by the pump manufacturer.

NEWTONIAN FLUID : A ﬂuid where the relation between shear stress and shear rate is linear, related to

viscosity.

NICKEL : A heavy metal commonly regulated by wastewater discharge permits. It can be found in met-

als–related industries and products, including stainless steel. Other heavy metals include – Arsenic (As),

Cadmium (Cd), Chromium(Cr), Copper (Cu), Lead (Pb), and Zinc (Zn).

NITRIFICATION : The conversion of nitrogen matter into nitrates by bacteria.

NITRIFICATION : An aerobic process in which bacteria change ammonia and organic nitrogen into

nitrite and nitrate forms of nitrogen.

NITRIFICATION : Wastewater treatment process involved in the biological removal of ammonia in which

the ammonia is converted to nitrates (NO3)

NITRIFYING BACTERIA : Bacteria that change the ammonia and organic nitrogen in wastewater into

oxidized nitrogen (usually nitrate).

NITROGEN : Nitrogen is present in wastewater in many forms – total Kjeldahl nitrogen, ammonia nitro-

gen, organic nitrogen.

NITROGEN CYCLE : The cycle of life, death, and decay involving organic nitrogenous matter is known

as the nitrogen cycle. In the nitrogen cycle ammonia is produced from proteins.

NON: Newtonian Fluid – A ﬂuid with properties that is different in any way from those of Newtonian

Fluids. This is usually found in the relation of viscosity and shear or shear/time.

NON: RISING STEM – A gate valve having its stem threaded into the gate. As the stem turns, the gate

moves, but the stem does not rise. Stem threads are exposed to line ﬂuids.

NON: Wetted Parts – A term used any part of a pump, or other type of liquid handling equipment, that

doesn’t come into direct contact with the process ﬂuid.

NONIONIC : Neutral charged high molecular weight polyelectrolyte water soluble organic polymer

designed to agglomerate solids in water substrates.

NONPOINT DISCHARGE : A source of wastewater that comes from a relatively large area and would

have to be controlled by a management or conservation practice. Storm waters and most agricultural

waters are nonpoint sources.

NONPOTABLE : Water that is considered unsafe and/or unpalatable for drinking.

NONTRANSIENT NONCOMMUNITY (NTNC) WATER SYSTEM : Means a public water system that

is not a community water system and that regularly serves at least 25 of the same persons over six months

per year, including schools, day care centers, factories, restaurants and hospitals. 232

NORMALLY CLOSED SOLENOID VALVE : An electrically operated valve whose inlet oriﬁce is closed

when the solenoid coil is not energized. Energize to open.

NPDES PERMIT : National Pollutant Discharge Elimination System permit is the regulatory agency doc-

ument issued by either a federal or state agency which is designed to control all discharges of pollutants

from all point sources and storm water runoff into U.S. waterways. A treatment plant that discharges to a

surface water will have a NPDES permit.

NRS : NON–RISING STEM – A gate valve having its stem threaded into the gate. As the stem turns the

gate moves but the stem does not rise. Stem threads are exposed to the line ﬂuid.

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Chapter 25. Glossary

NUTRIENTS : Substances required by living plants and organisms. Forms of nitrogen and phosphorous

are nutrients that can cause problems in receiving waters.

OBSTRUCTION : Any solid object in or protruding into a wastewater ﬂow in a collection line that pre-

vents a smooth or even passage of the wastewater.

OFFSET : (1) A combination of elbows or bends which brings one section of a line of pipe out of line

with, but into a line parallel with, another section. (2) A pipe ﬁtting in the approximate form of a reverse

curve, made to accomplish the same purpose. (3) A pipe joint that has lost its bedding support and one

of the pipe sections has dropped or slipped, thus creating a condition where the pipes no longer line up

properly.

OPDES : Oklahoma Pollutant Discharge Elimination System – A permit program established in accor-

dance with Section 402 of the CWA and authorized in 27A O.S. Environment and Natural Resources. This

program regulates discharges into Oklahoma’s waters from point sources, including municipal, industrial,

commercial and certain agricultural sources.

OPERATING POINT : The point on the system curve corresponding to the ﬂow and head required to

meet the process requirements.

OPERATOR : A device which converts manual, hydraulic, pneumatic or electrical energy into mechanical

motion to open and close a valve.

ORGANIC : Material which comes from mainly animal or plant sources and contains carbon.

ORGANIC ACIDS : Weak acids formed from organic compounds, such as acetic acid and citric acid.

These acids form ﬁrst in anaerobic digesters and then are converted to methane. The organic acids in

wastewater lagoons are much more complex and generally weaker.

ORGANIC MATERIAL : Carbon based material that can be broken down by bacteria (fats, meats, plant

life). Represents the volatile fraction of solids.

ORTHOPHOSPHATE : A simple compound of phosphorous and oxygen that is soluble in water.

OSANDY : OUTSIDE SCREW AND YOKE – A valve in which the ﬂuid does not come in contact with

the stem threads. The stem sealing elements is between the valve body and the stem threads.

OUTFALL : (1) The point, location or structure where wastewater or drainage discharges from a sewer,

drain, or other conduit. (2) The conduit leading to the ﬁnal disposal point or area.

OUTFALL SEWER : A sewer that receives wastewater from a collection system or from a wastewater

treatment plant and carries it to a point of ultimate or ﬁnal discharge in the environment.

OUTLET : Downstream opening or discharge end of a pipe, culvert, or canal.

OVERFLOW MANHOLE : A manhole which ﬁlls and allows raw wastewater to ﬂow out onto the street

or ground.

OVERFLOW RELIEF LINE : Where a system has overload conditions during peak ﬂows, an outlet may

be installed above the invert and leading to a less loaded manhole or part of the system. This is usually

called an “overﬂow relief line.”

OXIC : A biological environment which is aerobic.

OXIDATION : Oxidation is the addition of oxygen, removal of hydrogen, or the removal of electrons from

an element or compound. In wastewater treatment, organic matter is oxidized to more stable substances.

OXIDATION : Oxidation is the addition of oxygen, removal of hydrogen, or the removal of electrons from

511

an element or compound. In wastewater treatment, organic matter is oxidized to more stable substances.

OXIDATION POND : A term often used interchangeably with lagoon. Oxidation ponds are used after

other treatment processes.

OXIDATION PONDS OR LAGOONS : Holding ponds designed to allow the decomposition of organic

wastes by aerobic or anaerobic means.

OXIDATION PONDS OR LAGOONS : Holding ponds designed to allow the decomposition of organic

wastes by aerobic or anaerobic means.

OXYGEN REDUCTION POTENTIAL : A measure that indicates the capacity of wastewater to gain

or reduce electrons during a chemical reaction. It is used as a control parameter for treating hexavalent

chromium wastewater in the metal ﬁnishing industry.

PACKAGE TREATMENT PLANT : A small wastewater treatment plant often fabricated at the manufac-

turer’s factory, hauled to the site, and installed as one facility. The package may be either a small primary

or a secondary wastewater treatment plant.

PACKING : The deformable sealing material inserted into a valve stem stufﬁng box, which, when com-

pressed by a gland, provides a tight seal about the stem.

PALMER: BOWLUS FLUME – A ﬂow measuring device consisting of a preformed ﬂume.

PARACHUTE : A device used to catch wastewater ﬂow to pull a ﬂoat line between manholes.

PARSHALL FLUME : A ﬂow measuring device consisting of a preformed ﬂume with restrictive area

called the throat. The head of water at a stilling well just upstream from the narrow part of the throat is

measured and a chart is used to obtain ﬂow rate.

PARTS PER BILLION : A unit of concentration for pollutants in the wastewater. It’s the equivalent of one

microgram in 1 liter (ug/l).

PARTS PER MILLION : A unit of concentration for pollutants in the wastewater. It’s the equivalent of 1

milligram in 1 liter (mg/l).

PATHOGENIC ORGANISMS : bacteria, viruses or cysts, which can cause disease (typhoid, cholera,

dysentery) in a host such as a human. Also called Pathogens.

PEAKING FACTOR : Ratio of a maximum ﬂow to the average ﬂow, such as maximum hourly ﬂow or

maximum daily ﬂow to the average daily ﬂow.

PERCOLATION : The movement or ﬂow of waster through soil or rocks.

PERCOLATION : The movement or ﬂow of water through soil or rocks. A discharge option for many

wastewater treatment systems. In Montana, a Montana Ground Water Pollution Control System (MGW-

PCS) permit and sampling in groundwater monitoring wells are required for permitted systems.

PERFORMANCE CURVE : A curve of ﬂow vs. Total Head for a speciﬁc pump model and impeller

diameter.

PERSONAL COMPUTER (PC) : Any general–purpose computer whose size, capabilities, and original

sales price make it useful for individuals, and which is intended to be operated directly by an end–user

with no intervening computer operator.

PERSONAL PROTECTIVE EQUIPMENT (PPE) : Personal protective equipment refers to protective

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Chapter 25. Glossary

clothing, helmets, goggles or other garments or equipment designed to protect the wearer’s body from

injury by blunt impacts, electrical hazards, heat, chemicals and infection, for job–related occupational

safety and health purposes.

PH : The intensity of the basic or acid condition of a liquid.

PH VALUE : A convenient method of expressing small differences in the acidity or alkalinity of solutions.

Neutrality = pH 7.1; lower values indicate increasing acidity, higher values indicate increasing alkalinity.

PH VALUE : A convenient method of expressing small differences in the acidity or alkalinity of solutions.

Neutrality = pH 7.1; lower values indicate increasing acidity, higher values indicate increasing alkalinity.

PHOTOGRAPHIC INSPECTIONS : A method of obtaining photographs of a pipeline by pulling a

time–lapse motion picture camera through the line.

PHOTOSYNTHESIS : A complex process in all green plants that contain chlorophyll. The process uses

sunlight as energy to convert carbon dioxide into plant growth. As a by–product oxygen is released.

PIG : Refers to a poly pig which is a bullet–shaped device made of hard rubber or similar material.

PIPE CAPACITY : In a gravity–ﬂow sewer system, pipe capacity is the total amount in gallons a pipe is

able to pass in a speciﬁc time period.

PIPE CLEANING : Removing grease, grit, roots and other debris from a pipe run by means of one of the

hydraulic cleaning methods.

PIPE DIAMETER : The nominal or commercially designated inside diameter of a pipe, unless otherwise

stated.

PIPE DISPLACEMENT : The cubic inches of soil or water displaced by one foot or one section of pipe.

PIPE FRICTION LOSS : The positive head loss from the friction resistance between the pipe walls and

the moving liquid.

PIPE GRADE : The angle of a sewer or a single section of a sewer as installed. Usually expressed in a

percentage ﬁgure to indicate the drop in feet or tenths of a foot per hundred feet. For example, 0.5 percent

grade means a drop of one–half foot per 100 feet of length.

PIPE JOINT : A place where two sections of pipe are coupled or joined together.

PIPE JOINT SEAL : (1) The tightness or lack of leakage at a pipe joint. (2) The method of sealing a pipe

coupling.

PIPE LINER : A plastic liner pulled or pushed into a pipe to eliminate excessive inﬁltration or exﬁltration.

Other solutions to the problem of inﬁltration/exﬁltration are the use of cement grouting or replacement of

damaged pipe.

PIPE PLUG : (1) A temporary plug placed in a sewer pipe to stop a ﬂow while repair work is being ac-

complished or other functions are performed. (2) In construction of a new sewer system, service saddles

are sometimes installed before a building or a building lateral is in existence. Under such circumstances, a

plug will be placed in the off–lead of the saddle of a “Y.”

PIPE RODDING : A method of opening a plugged or blocked pipe by pushing a steel rod or snake, or

pulling same, through the pipe with a tool attached to the end of the rod or snake. Rotating the rod or

snake with a tool attached increases effectiveness. 234

PIPE ROUGHNESS : A measurement of the average height of peaks producing roughness on the internal

surface of pipes. Roughness is measured in many locations, and is usually deﬁned in micro–inches RMS

513

(root mean square).

PIPE RUN : (1) The length of sewer pipe reaching from one manhole to the next. (2) Any length of pipe,

generally assumed to be in a straight line.

PIPE SECTION : A single length of pipe between two joints or couplers.

PIPING & INSTRUMENTATION DIAGRAM (P&ID) : A diagram in the process industry that shows the

piping of the process ﬂow together with the installed equipment and instrumentation.

PISTON EFFECT : The sealing principle involved in utilizing line pressure to effect a seal across the

ﬂoating seats of some valves.

PISTON PUMP : The piston pump is a positive displacement type of pump. As the piston is pulled back

it draws in the ﬂuid, and then as it’s pushed forward it pushes the liquid out. A piston pump can have up

to four pistons depending on the application. They should only be used for clear liquids as any solids

and/or abrasives in the ﬂuid can damage the pump. Piston pumps are for low ﬂow, high head applications.

Frequently used for high–accuracy metering applications.

PLAN : A drawing showing the TOP view of sewers, manholes and streets. Also means approved contract

drawings, town standards, working drawings, detail sheets or exact reproductions thereof, which show the

location, character, dimensions and details of the work to be done.

PLUG : The rotating closure element of a plug valve. Also a threaded ﬁtting used to close off and seal an

opening into a pressure containing chamber, e.g., pipe plug.

PLUG VALVE : A quarter turn valve whose closure element is usually a tapered plug having a rectangular

port.

PNEUMATIC EJECTOR : A device for raising wastewater, sludge or other liquid by compressed air.

The liquid is alternately admitted through an inward–swinging check valve into the bottom of an airtight

pot. When the pot is ﬁlled compressed air is applied to the top of the liquid. The compressed air forces

the inlet valve closed and forces the liquid in the pot through an outward–swinging check valve, thus

emptying the pot.

POLISHING POND : A ﬁnal lagoon cell after other treatment which completes the treatment, or "pol-

ishes" the efﬂuent.

POLY PAK STEM SEAL : An O–ring energized lip–seal which replaces O–ring stem seals in certain gate

valves. Also used for stem seals in some ball valves.

POLYELECTROLYTES : Synthetic chemicals used as a coagulant aid.Primary Waste Treatment – Me-

chanical separation of solids, grease, and scum from wastewater. With the aid of ﬂocculating agents,

primary treatment can eliminate 50 to 65 POLYMER : Polymers are used with other chemical coagulants

to aid in binding small suspended particles to larger chemical ﬂocs for their removal from water.

POLYMER : Polymers are used with other chemical coagulants to aid in binding small suspended parti-

cles to larger chemical ﬂocs for their removal from water.

POLYPHOSPHATE : A large compound formed of several orthophosphate molecules connected by

phosphate–storing microorganisms.

POLYVINYL CHLORIDE : The most common material used for wastewater piping. It is a type of plas-

tic.

PONDING : A condition occurring on trickling ﬁlters when the hollow spaces (voids) become plugged to

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Chapter 25. Glossary

the extent that water passage through the ﬁlter is inadequate. Ponding may be the result of excessive slime

growths, trash, or media breakdown.

PONDING : A condition occurring on trickling ﬁlters when the hollow spaces (voids) become plugged to

the extent that water passage through the ﬁlter is inadequate. Ponding may be the result of excessive slime

growths, trash, or media breakdown.

POPULATION EQUIVALENT (HYDRAULIC) : A ﬂow of 100 gallons per day is the hydraulic or ﬂow

equivalent to the contribution or ﬂow from one person. Population equivalent = 100 GPCD or gallons per

capita per day.

POPULATION EQUIVALENT : An average BOD contribution by each person to a domestic sewage. The

accepted population equivalent is 0.17 pounds of BOD per person per day.

PORCUPINE : A sewer cleaning tool the same diameter as the pipe being cleaned. The tool is a steel

cylinder having solid ends with eyes cast in them to which a cable can be attached and pulled by a winch.

Many short pieces of cable or bristles protrude from the cylinder to form a round brush.

POTABLE WATER : Water ﬁt for human consumption.

POTENTIAL ENERGY : A thermodynamic property. The energy associated with the mass and height of

a body above a reference plane.

POTENTIAL HYDROGEN : A measurement of how acidic or basic wastewater is on a scale of 0 to 14.

POUNDS PER SQUARE INCH : A measurement of pressure. It’s often used when discussing physical

wastewater treatment technologies involving ﬁltration, but is also used with pumps. Filtration system PSI

can indicate when it’s time to backwash or change a ﬁlter.

POWDER PUMP : Powder pumps are normally of the Air Operated Diaphragm type and really can pump

just powder and powder like materials such as ﬂour and other ﬁne grained, low bulk, dry–density powders

in a dust free operation.

POWER OPERATOR : Powered valve operators are of the following general types.. Electric Motor,

Pneumatic or Hydraulic Motor, Pheumatic or Hydraulic Cylinder. Operators can either be adapted directly

to the valve stem or side mounted on existing gear or scotch–yoke operators.

POWER RODDER : A sewer cleaning machine ﬁtted with auger rods which are inserted in a sewer line to

dislodge and cut roots and debris.

PPM : Abbreviation for Parts Per Million. See MILLIGRAMS PER LITER (mg/L)

PRE: CLEANING – Sewer line cleaning, commonly done by high–velocity cleaners, that is done prior to

the TV inspection of a pipeline to remove grease, slime, and grit to allow for a clearer and more accurate

identiﬁcation of defects and problems. 235

PRECIPITATE : (1) An insoluble, ﬁnely divided substance which is a product of a chemical reaction

within a liquid. (2) The separation from solution of an insoluble substance.

PRECIPITATION : (1) The total measurable supply of water received directly from clouds as rain, snow,

hail, or sleet; usually expressed as depth in a day, month, or year, and designated as daily, monthly, or

annual precipitation. (2) The process by which atmospheric discharged onto a land or water surfaces. (3)

The separation (of a substance) out in solid form from a solution, as by the use of a reagent. moisture is

PRELIMINARY TREATMENT : The removal of rocks, rags, sand, eggshells, and similar materials which

may hinder the operation of a treatment plant. Preliminary treatment is accomplished by using equipment

515

such as bar screens and grit removal systems.

PRESSURE : The application of external or internal forces to a body producing tension or compression

within the body. This tension divided by a surface is called pressure.

PRESSURE DROP : Referring to the loss of pressure between two points in a pipeline system. Generally,

this occurs because of pipe friction loss or differences in elevation between the two points.

PREVENTIVE MAINTENANCE : Crews assigned the task of cleaning sewers (for example, balling

or high–velocity cleaning crews) to prevent stoppages and odor complaints. Preventive maintenance is

performing the most effective cleaning procedure, in the area where it is most needed, at the proper time

in order to prevent failures and emergency situations.

PRIMARY CELL : The ﬁrst cell in a series, generally receiving raw wastewater.

PRIMARY CONTAMINANTS : The contaminants identiﬁed by the EPA as harmful to human health. In

order to protect public health, the primary contaminants must not exceed certain speciﬁed levels known as

Maximum Contaminant Levels (MCL).

PRIMARY TREATMENT (ALSO KNOWN AS SEDIMENTATION) : A wastewater treatment process

that takes place in a rectangular or circular tank and allows those substances in wastewater that readily

settle or ﬂoat o be separated from the water being treated.

PRIMARY WASTE TREATMENT : Mechanical separation of solids, grease, and scum from wastewater.

With the aid of ﬂocculating agents, primary treatment can eliminate 50 PROCESS AND INSTRUMEN-

TATION DIAGRAM : An engineering drawing for a wastewater treatment system. It’s a schematic ﬂow

diagram that shows the relationship between different instrumentation and equipment.

PROCESS CONTROL : An engineering discipline that deals with architectures, mechanisms and algo-

rithms for maintaining the output of a speciﬁc process within a desired range.

PROCESS FLOW DIAGRAM : A diagram commonly used in engineering to indicate the general ﬂow of

plant processes and equipment. The PFD displays the relationship between major equipment of a plant

facility and does not show minor details.

PROCESS IN WHICH CAVITIES OR BUBBLES FORM IN THE FLUID LOW: pressure area and col-

lapse in a higher pressure area of the pump – causing noise, damage to the pump, and loss of efﬁciency

because it distorts the ﬂow pattern. Occurs in centrifugal pumps when NPSHa < NPSH. A properly de-

signed system and a properly sized pump will prevent cavitation.

PROFILE : A drawing showing the SIDE view of sewers and manholes.

PROGRAMMABLE LOGIC CONTROLLER (PLC) : A microprocessor–based system which provides

plant automation by monitoring sensors and controlling actuators and equipment in real time.

PROGRESSIVE CAVITY PUMP : A pump that uses a stator and rotor in a screw shape. The rotor turning

inside the stator causes cavities to move in the direction of ﬂow. Progressive Cavity Pumps cannot run dry

for any amount of time.

PROOF PPRESSURE : A hydrostatic test pressure, usually 1 ½ times the rated working pressure, applied

to an assembled valve to verify the structural integrity of the pressure containing parts. Synonymous with

hydrostatic shell test. (Table 5.1, API–6D).

PROPELLER PUMP : Propeller pumps are similar to other centrifugal impeller pumps, but the ﬂuid being

pumped is not sent in a circular path. Rather, it proceeds more or less in a straight direction up to the

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Chapter 25. Glossary

discharge. The motor sits above the discharge shaft. The propeller can be placed below the surface of the

liquid, where it will always be primed. Propeller pumps are generally low–speed but low heads. They can

be quite large, measuring over a dozen feet in diameter and moving over 50,000 gallons per minute. Some

have adjustable–pitch blades.

PROTECTIVE SLEEVES : A circular "pipe like" sleeve inserted in place of the ball and seats of a top

entry ball valve. This protective sleeve remains in place inside the valve during valve installation and

ultimate pigging of a pipeline to clear debris from the line before placing the pipeline into service. Once

the pipeline has been purged of all debris, the protective sleeve is removed entirely from the ball valve

cavity and operating trim (i.e. ball and seats) is then installed for normal service conditions.

PUBLIC SEWER : means a sewer in which all owners of abutting properties have equal rights and is

controlled by acting as Sewer Commissioners, and maintained by the Public Works Superintendent.

PUBLICLY OWNED TREATMENT WORK : A term to describe a city or municipal sewage treatment

facility. Since most industries discharge wastewater to these facilities, they’re typically regulated by these

POTWs.

PUMP : A mechanical device for causing ﬂow, for raising or lifting water or other ﬂuid, or for applying

pressure to ﬂuids.

PUMP CONTROL VALVE : A ball valve that is not meant for on–off service, but whose speciﬁc function

is to control ﬂow and prevent cavitation in pumps on liquid pipelines.

PUMP IMPELLER : The moving element in a centrifugal pump that drives the ﬂuid.

PUMP PIT : A dry well, chamber or room below ground level in which a pump is located.

PUMP STATION : Installation of pumps to lift wastewater to a higher elevation in places where ﬂat land

would require excessively deep sewer trenches. Also used to raise wastewater from areas too low to drain

into available collection lines. These stations may be equipped with

RACHET DRIVE : A shaft or valve that is operated by means of a ratchet mechanism. The ratchet deliv-

ers an intermittent stepped rotation through a gear in one direction only.

RAW WASTEWATER : Wastewater before it receives any treatment.

REACTOR : A tank where a wastewater stream is mixed with bacterial sludge and biochemical reactions

occur.

RECEIVING WATER : A stream, river, lake, ocean or other surface or groundwater into which treated or

untreated wastewater is discharged.

RECIRCULATION : The return of part of the efﬂuent from a treatment process to the incoming ﬂow.

REDUCED PORT : A valve port opening that is smaller than the line size or the valve end connection

size.

REGULAR PORT VALVE : A term usually applied to plug valves. The "regular" port of such a valve

is customarily about 40 REGULATOR : A throttling valve which exercises automatic control over some

variable (usually pressure). Not an on–off valve.

RELIEF VALVE : A quick acting, spring loaded valve that opens (relieves) when the pressure exceeds the

spring setting. Often installed on the body cavity of ball and gate valves to relieve thermal overpressure in

liquid services.

REMOTE CONTROL : The operation of a valve or other ﬂow control device from a point at a distance

517

from the device being controlled. Can be accomplished by electrical, pneumatic or hydraulic means.

RESILIENT SEAT : A valve seat containing a soft seal, such as an o–ring, to assure tight shut–off.

RETENTION : (1) That part of the precipitation falling on a drainage area which does not escape as

surface stream ﬂow during a given period. It is the difference between total precipitation and total runoff

during the period, and represents evaporation, transpiration, subsurface leakage, inﬁltration, and when

short periods are considered, temporary surface or underground storage on the area. (2) The delay or

holding of the ﬂow of water and water carried wastes in a pipe system. This can be due to a restriction in

the pipe, a stoppage or a dip. Also, the time water is held or stored in a basin or wet well.

RETENTION TIME : The time water, sludge or solids are retained or held in a clariﬁer or sedimentation

tank.

RETURN ACTIVATED SLUDGE : Activated return sludge is normally returned continuously to the

aeration tank. Recycling of activated sludge back to the aeration tank provides bacteria for incoming

wastewater. Its should be brown in color with no obnoxious odor and is often also returned in small por-

tions to the primary settling tanks to aid sedimentation. Settled activated sludge is generally thinner than

raw sludge. Some activated sludge will be wasted to prevent excessive solids build up.

RETURN ACTIVATED SLUDGE SOLIDS (RASS) : The concentration of suspended solids in the sludge

ﬂow being returned from the settling tank to head of the aeration tank.

RETURN SLUDGE : Settled activated sludge returned to mix with incoming raw or primary settled

wastewater. When the return sludge rate in the activated sludge process is too low, there will be insufﬁ-

cient organisms to meet the waste load entering the aerator.

REVERSE OSMOSIS : Water puriﬁcation technology that uses a semipermeable membrane to remove

dissolved solids, molecules and larger particles from water. Applied pressure is used to overcome osmotic

pressure and produce high purity water. Reverse osmosis is used to produce ultra pure water for a variety

of applications.

RIM PULL : The force required at the edge of the handwheel to generate the required torque at the center

of the handwheel.

RIP: RAP – Erosion control by placement of large rocks along an embankment.

RISING SLUDGE : Rising sludge occurs in the secondary clariﬁers of activated sludge plants when the

sludge settles to the bottom of the clariﬁer, is compacted, and then starts to rise to the surface, usually as a

result of denitriﬁcation.

RISING STEM : A valve stem which rises as the valve is opened. A valve stem with threads arranged

so that as the stem turns, the threads engage a stationary threaded area and lift the stem along with the

closure element attached to it.

RISING STEM BALL VALVE : A single seated ball valve that is designed to seal by using the valves

stem to mechanically wedge the valves ball into a stationary seat effecting a bubble tight seal. The valves

stem operates througha guide sleeve assembly that guides the stem through a quarter turn of rotation as the

stem is raised or lowered by a handwheel (or actuator). The mechanical action of the stem moves the ball

away fromthe seat prior to the 90° rotation of the ball. This design provides lower operating torques and

longer seat life while assuring bubble tight shut off.

ROD (SEWER) : A light metal rod, three to ﬁve feet long with a coupling at each end. Rods are joined

and pushed into a sewer to dislodge obstructions.

518

Chapter 25. Glossary

ROD GUIDE : A bent pipe inserted in a manhole to guide hand and power rods into collection lines so the

rods can dislodge obstructions.

RODDING MACHINE : A machine designed to feed a rod into a pipe while rotating the rod. 236

RODDING TOOLS : Special tools attached to the end of a rod or snake to accomplish various results in

pipe rodding.

ROOF LEADER : A downspout or pipe installed to drain a roof gutter to a storm drain or other means of

disposal.

ROOT MOP : When roots from plant life enter a sewer system, the roots frequently branch to form a

growth that resembles a string mop.

ROOT SEWER : Any part of a root system of a plant or tree that enters a collection system.

ROTARY GEAR PUMP : A Gear pump uses the meshing of gears to pump ﬂuid by displacement. They

are one of the most common types of pumps for hydraulic ﬂuid power applications. Gear pumps are also

widely used in chemical installations to pump ﬂuid with a certain viscosity. There are two main variations;

external gear pumps which use two external spur gears and internal gear pumps which use an external and

an internal spur gear. Gear pumps are ﬁxed displacement, meaning they pump a constant amount of ﬂuid

for each revolution. Some gear pumps are designed to function as either a motor or pump.

ROTATING BIOLOGICAL CONTACTOR : A biological treatment technology most often used in city

treatment systems to reduce BOD.

ROTIFER : A form of microscopic animal that feeds on algae and bacteria. The free swimming protozoa

are common in lagoons. Rotifers require aerobic conditions.

SADDLE : A ﬁtting mounted on a pipe for attaching a new connection. This device makes a tight seal

against the main pipe by use of a clamp, adhesive, or gasket and prevents the service pipe from protruding

into the main.

SADDLE CONNECTION : A building service connection made to a sewer main with a device called a

saddle.

SAFETY VALVE : A quick opening, pop action valve used for fast relief of excessive pressure.

SAND TRAP : A device which can be placed in the outlet of a manhole to cause a settling pond to de-

velop in the manhole invert, thus trapping sand, rocks and similar debris heavier than water. Also may be

installed in outlets from car wash areas.

SANITARY COLLECTION SYSTEM : The pipe system for collecting and carrying liquid and liq-

uid–carried wastes from domestic sources to a wastewater treatment plant.

SANITARY PUMP : Sanitary pumps describe the materials used for construction of how a pump is built

and if they meet speciﬁc criteria set forth by certifying agencies. Typical describing words are “FDA

Compliant”, “Food Grade” and “CIP (Clean in Place)” and EHEDG. Sanitary pumps are normally built

from stainless steel, PTFE, EPDM and other “clean” materials.

SANITARY SEWER : A pipe or conduit (sewer) intended to carry wastewater or waterborne wastes

from homes, businesses, and industries to the POTW. Storm water runoff or unpolluted water should be

collected and transported in a separate system of pipes or conduits (storm sewers) to natural water courses.

SATURATION : Oxygen saturation is the concentration of free dissolved oxygen in water that is in equi-

librium with atmospheric oxygen. It is measured in milligrams per liter (mg/l). It varies with both temper-

519

ature and atmospheric pressure.

SCOOTER : A sewer cleaning tool whose cleansing action depends on the development of high water

velocity around the outside edge of a circular shield. The metal shield is rimmed with a rubber coating

and is attached to a framework on wheels (like a child’s scooter). The angle of the shield is controlled by a

chain–spring system which regulates the head of water behind the scooter and thus the cleansing velocity

of the water ﬂowing around the shield.

SCOTCH YOKE OPERATOR (USED ON QUARTER TURN VALVES) : A quarter turn operator using

a scotch yoke mechanism rather than gears. The "Scotch Yoke" has a torque output at the beginning and

ending of its stroke that is generally twice the magnitude oft he torque output in the center of its stroke.

SCREEN : A device used to retain or remove suspended or ﬂoating objects in wastewater. The screen

has openings that are generally uniform in size. It retains or removes objects larger than the openings. A

screen may consist of bars, rods, wires, gratings, wire mesh, or perforated plates.

SCREEN : A device used to retain or remove suspended or ﬂoating objects in wastewater. The screen

has openings that are generally uniform in size. It retains or removes objects larger than the openings. A

screen may consist of bars, rods, wires, gratings, wire mesh, or perforated plates.

SCREW CENTRIFUGAL PUMP : A pump which uses an open channel impeller with a screw shape.

These pumps are ideal for sludges, large/stringy solids laden ﬂuids, shear sensitive ﬂuids and delicate or

highly abrasive materials. Offering true non–clog performance and a steep head curve make these pumps

ideal for wastewater and other sludge applications. They also handle solids more gently than other pumps.

They are speciﬁcally used for transporting live ﬁsh without harm and delicate foodstuffs without bruising.

SCUM : (1) A layer or ﬁlm of foreign matter (such as grease, oil) that has risen to the surface of water or

wastewater. (2) A residue deposited on the ledge of a sewer, channel, or wet well at the water surface. (3)

A mass of solid matter that ﬂoats on the surface.

SEAT : That part of a valve against which the closure element (gate, ball) effects a tight shut–off. In many

ball valves and gate valves, it is a ﬂoating member containing a soft seating element (usually an o–ring).

SECONDARY : The second in a series of cells.

SECONDARY CONTAMINANTS : Contaminants in drinking water that are not harmful to human health

but are unpleasant. Secondary contaminants include substances that cause unpleasant tastes and odors

or color the water. A Recommended Maximum Level (RCM) has been set for each of the secondary

contaminants in order to make sure the water is pleasant to drink. 237

SECONDARY TREATMENT : A wastewater treatment process used to convert dissolved or suspended

materials into a form more readily separated from wastewater. Usually follows primary sedimentation

treatment and uses biological processes to convert wastes to solids that settle in secondary clariﬁers. Also

occurs in lagoon systems.

SECONDARY WASTE TREATMENT : Processing by various types of systems that employ aeration and

biological oxidation stages to decompose dissolved and colloidal organic contaminants (inorganic plant

nutrients may also be partially removed).

SEDIMENT : Solid material settled from suspension in a liquid.

SEDIMENTATION : The process of settling and depositing of suspended matter carried by wastewater.

Sedimentation usually occurs by gravity when the velocity of the wastewater is reduced below the point at

which it can transport the suspended material.

520

Chapter 25. Glossary

SEDIMENTATION TANKS : Provide a period of quiescence during which suspended waste material

settles to the bottom of the tank and is scraped into a hopper and pumped out for disposal. During this

period, ﬂoatable solids (fats, oils) rise to the surface of te tank and are skimmed off into scum pipes for

disposal.

SELECT BACKFILL : Material used in backﬁlling of an excavation, selected for desirable compaction or

other characteristics.

SELECT BEDDING : Material used to provide a bedding or foundation for pipes or other underground

structures. This material is of speciﬁed quality for desirable bedding or other characteristics and is often

imported from a different location.

SELF RELIEVING : The process whereby excessive internal body pressure, in some valves, is automat-

ically relieved either into the upstream or downstream line by forcing the seats away from the closure

element.

SELF: Priming Pump – Self–priming pumps are centrifugal pumps with an abnormally large and specially

shaped volute. The purpose of the large volute is to allow the pump to pull or “lift” liquid up to the im-

peller. Initially the pump volute (casing), must be ﬁlled with liquid manually to “pre–prime” the pump. As

the pump starts it pumps out the liquid that was manually put into it while also drawing up the air in the

suction pipe along with pulling up the liquid to be pumped. As the lifted liquid enters the volute the ﬁnal

volume of air is pumped out of the discharge and through an air release valve. Once the liquid hits the air

valve, it closes and the pump now operates as a standard centrifugal pump.

SEPTIC : A condition that exists when there is no dissolved oxygen (see anaerobic). Anaerobic bacteria

and other microorganisms continue to use parts of the waste for food, but produce foul odors and black

colored water. The waste in the common septic tank is typical of this condition.

SEQUENCING BATCH REACTORS : A biological treatment technology based on the activated sludge

process. It is sometimes used in small municipalities and at food processing facilities who discharge

directly to streams or rivers.

SERVICE ROOT : A root entering the sewer system in a service line and growing down the pipe and into

the sewer main.

SETTLEABILITY : A process–control test used to evaluate the settling characteristics of activated sludge.

Reading taken at 30 to 60 minutes are used to calculate the settled sludge volume (SSV) and the sludge

volume index (SVI).

SETTLED SLUDGE VOLUME (SSV) : The volume in percent occupied by an activated sludge sample

after 30 to 60 minutes of settling. Normally written as SSV with a subscript to indicate the time of the

reading used for calculation (SSV60) or (SSV30).

SEWAGE : Largely the water supply from a community after it has been fouled by various uses. From

the standpoint of course, it may be a combination of the liquid or water–carried wastes from residences,

business buildings, and institutions, together with those from industrial establishments, and with such

ground water, surface water, and storm water as may be present.

SEWER : A pipe or conduit that carries wastewater or drainage water.

SEWER BALL : A spirally grooved, inﬂatable, semi–hard rubber ball designed for hydraulic cleaning of

sewer pipes.

SEWER CLEANOUT : A capped opening in a sewer main that allows access to the pipes for rodding and

521

cleaning. Usually such cleanouts are located at terminal pipe ends or beyond terminal manholes.

SEWER GAS : (1) Gas in collection lines (sewers) that results from the decomposition of organic matter

in the wastewater. When testing for gases found in sewers, test for lack of oxygen and also for explosive

and toxic gases. (2) Any gas present in the wastewater collection system, even though it is from such

sources as gas mains, gasoline, and cleaning ﬂuid.

SEWER JACK : A device placed in manholes which supports a yoke or pulley that keeps wires or cables

from rubbing against the inlet or outlet of a sewer.

SEWER MAIN : A sewer pipe to which building laterals are connected.

SEWER USE DISCHARGE PERMIT : Permit required or issued jointly by the Authority and a Munici-

pality for the discharge of industrial waste. 238

SEWERAGE : System of piping with appurtenances for collecting, moving and treating wastewater from

source to discharge.

SEWERAGE SYSTEM : Any device, equipment or works used in the transportation, pumping, storage,

treatment, recycling, and reclamation of Wastewater and Industrial Wastes.

SEWERS : A system of pipes used for collecting domestic and industrial waste, as well as storm water

run–off. Lateral sewers connect homes and industries to trunk sewers, which channel waste into intercep-

tor sewers carry only domestic and industrial wastewater. Storm sewers carry only storm water run–off.

Combined sewers carry both.

SHORING : Material such as boards, planks or plates, and jacks used to hold back soil around trenches

and to protect workers in a trench from cave–ins.

SHORT CIRCUITING : A condition that occurs in tanks or basins when some of the water travels faster

than the rest of the ﬂowing water. This is usually undesirable since it may result in shorter contact, reac-

tion, or settling times in comparison with the theoretical (calculated) or presumed detention times.

SHORT GATE : A gate valve whose seat rings contact the gate only in the closed position. Such valves

are not through conduit, as the gate is completely withdrawn from the ﬂow area in the open position.

SHORT PATTERN VALVE : A valve whose face–to–face dimension is less than the API–6D standard.

SHUT: off Head – The Total Head corresponding to zero ﬂow on the pump performance curve.

SHUT: OFF VALVE – A valve designed only for on/off service. Not a throttling valve. Sometimes re-

ferred to as a "block valve."

SILT DENSITY INDEX : Measurement of silt, colloids, bacteria and other foulants of Reverse Osmo-

sis (RO) membranes. SDI is used to help determine the suitability of water or other liquids for the RO

process.

SILTING : Silting takes place when the pressure of inﬁltrating waters is great enough to carry silt, sand

and other small particles from the soil into the sewer system. Where lower velocities are present in the

sewer pipes, settling of these materials results in silting of the sewer system.

SILVER : A metal element regulated by wastewater discharge permits and common in metal ﬁnishing

wastewater.

SIPHON : Is a system of piping or tubing where the exit point is lower than the entry point.

SLAB GATE : A gate having ﬂat, ﬁnely ﬁnished, parallel faces – as opposed to a wedge gate. Such a

closure element slides across the seats and does not depend on stem force to achieve tight shut off.

522

Chapter 25. Glossary

SLAM RETARDER : A device designed to prevent the clapper of a check valve from slamming as it

closes upon ﬂow reversal. Hydraulic damping cylinders, rotary vanes, and torsional springs are all used

for this purpose.

SLEEVE : A pipe ﬁtting for joining two pipes of the same nominal diameter in a straight line.

SLIPLINING : A sewer rehabilitation technique accomplished by inserting ﬂexible polyethylene pipe into

an existing deteriorated sewer.

SLOPE : The slope or inclination of a sewer trench excavation is the ratio of the vertical distance to the

horizontal distance or “rise over run.” The inclination of a trench bottom or a trench sidewall, expressed as

a ratio of vertical distance to the horizontal distance. For example, a 3:1 slope shall rise or fall 3’ vertical

feet in a distance of 1’ horizontal foot.

SLUDGE : The accumulated suspended solids of sewage deposited in tanks or basins.

SLUDGE : (1) The settleable solids separated from liquids during processing. (2) The deposits of foreign

material on the bottoms of streams or other bodies of water.

SLUDGE AGE : In the activated sludge process, a measure of the length of time a particle of suspended

solids has been undergoing aeration, expressed in day. It is usually computed by dividing the weight of the

suspended solids in the aeration tank by the weight of excess activated sludge discharged from the system

per day.

SLUDGE DIGESTION : The process of changing organic matter in sludge into a gas or liquid or a more

stable solid form. These changes take place as microorganisms feed on sludge in anaerobic (more com-

mon) or aerobic digesters.

SLUDGE DIGESTION : The process of changing organic matter in sludge into a gas or liquid or a more

stable solid form. These changes take place as microorganisms feed on sludge in anaerobic (more com-

mon) or aerobic digesters.

SLUDGE DIGESTION : The purpose of sludge digestion is to separate the liquid from the solids to

facilitate drying. The proper pH range for digested sludge is 6.8 – 7.2.

SLUDGE INDEX : Properly called sludge volume index (SVI). It is the volume in millimeters occupied

by 1 g of activated sludge after settling of the aerated liquid for 30 minutes.

SLUDGE JUDGE : A clear tubular device used to measure sludge depth in clariﬁers or other tanks.

SLUDGE LOADING RATE : The weight of wet bio–solids fed to the reactor per square foot of reactor

bed area per hour (lb./ft2/H).

SLUDGE REAERATION : The continuous aeration of sludge after initial aeration for the purpose of

improving or maintaining its condition.

SLUDGE VOLUME INDEX (SVI) : A process–control calculation used to evaluate the settling quality of

activated sludge. Requires SSV30 and mixed liquor suspended solids test results to calculate.

SLURRY : Slurry is deﬁned as a suspension of solids in a liquid. Typically, the liquid is water and the

solids can be anything from soft materials such as sewage and food processing waste (potato skins, ﬁsh

parts) to abrasive solids like sand, ﬂy ash and coal. Keep in mind a typical centrifugal pump really can’t

handle more than 3 SMOKE TEST : A method of blowing smoke into a closed–off section of a sewer

system to locate sources of surface inﬂow.

523

SNAKE : A stiff but ﬂexible cable that is inserted into sewers to clear stoppages.

SOAP CAKE OR SOAP BUILDUP : A combination of detergents and greases that accumulate in sewer

systems, build up over a period of time, and may cause severe ﬂow restrictions.

SOIL POLLUTION : The leakage (exﬁltration) of raw wastewater into the soil or ground area around a

sewer pipe.

SOLENOID VALVE : A small electrically operated valve used in the control piping of powered by hy-

draulic or pneumatic cylinder operators.

SOLIDS FEED RATE : The dry solids fed to a centrifuge.

SOLIDS HANDLING PUMP : Many types of pumps can be used for solids handling. The size and con-

centration of solids in the ﬂuid will determine the best type of pump for the application. For sewage

applications such as lift stations where the solids concentration does not exceed 3 SOLIDS LOADING

(BELT FILTER PRESS) : The feed solids to the belt ﬁlter on a dry weight basis including chemicals per

unit time.

SOLIDS LOADING RATE (DRYING BEDS) : The weight of solids on a dry weight basis applied annu-

ally per square foot of drying bed area.

SOLIDS RECOVERY (CENTRIFUGE) : The ratio of cake solids to feed solids for equal sampling times.

It can be calculated with suspended solids and ﬂow data or with only suspended solids data. The cenrate

solids must be corrected if chemicals are fed to the centrifuge.

SOLUBLE BOD : Soluble BOD is the BOD of water that has been ﬁltered in the standard suspended

solids test.

SOLUTION : A liquid mixture of dissolved substances. In a solution it is impossible to see all the sepa-

rated parts.

SOUNDING ROD : A T–shaped tool or shaft that is pushed or driven down through the soil to locate

underground pipes and utility conduits.

SPECIFIC GRAVITY : The ratio of the density of a ﬂuid to that of water at standard conditions

SPECIFIC SPEED : A formula that describes the shape of a pump impeller. The higher the speciﬁc speed

the less N.P.S.H. required.

SPLIT CASE PUMP : The split case is almost synonymous with multi–stage and can be either horizontal

or vertical. Used where high pressures are needed such as boiler feed.

SPLITTER BOX : A division box that splits the incoming ﬂow into two or more streams. A device for

splitting and directing discharge from the head box to two separate points of application.

SPOIL : Excavated material such as soil from the trench of a sewer.

SPST : SINGLE–POLE, SINGLE THROW – Refers to the function of an electrical switch often used in

the control system of electric valve operators.

SPUR GEAR : The simplest of gears. In a gear set, the input spur gear and output spur gear are aligned

on parallel shafts. An idler gear may be used to the direction of rotation on the two shafts is in the same

direction.

SQUARE OPERATING NUT : A nut, usually 2" x 2", which is attached to a valve stem or the pinion

shaft of a gear operator allowing use of wrenches to quickly operate the valve.

524

Chapter 25. Glossary

SSIV (SUB SEA ISOLATION VALVE) : A valve used underwater, generally in a manifold that will close

and isolate a particular pipeline or process in an emergency.

STABILIZATION : The conversion of biodegradable materials into more stable solids. Stabilization is

the primary function of wastewater lagoons and treatment plants. Lagoons are often called stabilization

ponds.

STATION : A point of reference or location in a pipeline is sometimes called a “station.” As an example, a

building service is located 51 feet downstream from a manhole could be reported to be at “station 51.”

STEM : A rod or shaft used to transmit motion from an operator to the closure element of a valve.

STEM INDICATOR (VPI : VISIBLE POSITION INDICATOR) – A position indicating rod supplied with

gate valves. It extends from the top of the valve stem and serves to indicate the relative position of the

gate.

STEM NUT : A one or two–piece nut which engages the stem threads of a valve and transmits torque

from an operator to the valve stem.

STILLING WELL : A well or chamber which is connected to the main ﬂow channel by a small inlet.

Waves and surges in the main ﬂow stream will not appear in the well due to the small diameter inlet. The

liquid surface in the well will be quiet, but will follow all of the steady ﬂuctuations of the open channel.

The liquid level in the well is measured to determine the ﬂow in the main channel.

STOP COLLAR : The collar on a ball valve which restricts the ball to 90° of rotation from the fully open

to the fully closed position.

STOPPAGE : Partial or complete interruption of ﬂow as a result of some obstruction in a sewer. (2) When

a sewer system becomes plugged and the ﬂow backs up, it is said to have a “stoppage.”

STORM COLLECTION SYSTEM : A system of gutters, catch basins, yard drains, culverts and pipes

for the purpose of conducting storm waters from an area, but intended to exclude domestic and industrial

wastes.

STORM SEWER : A separate pipe, conduit or open channel (sewer) that carries runoff from storms,

surface drainage, and street wash, but does not include domestic and industrial wastes.

STRAIN : The ratio between the absolute displacements of a reference point within a body to a character-

istic length of the body.

STRATIFICATION : The formation of indistinct layers of slightly variable density of waters. Often

caused by warming of the surface with an absence of mixing.

STRESS : In this case refers to tangential stress or the force between the layers of ﬂuid divided by the

surface area between the layers.

STRETCH : Length of sewer from manhole to manhole. 240

STUFFING BOX : The annular chamber provided around a valve stem in a sealing system into which

deformable packing is introduced.

SUBMERSIBLE PUMP : Just like it sounds, these guys operate within the ﬂuid they are pumping. Sub-

mersible pumps can be either centrifugal or AOD type pumps. The centrifugal versions are common used

in sewage lift stations, while the AODs are used in chemical transfers.

SUCKER RODS : Rigid, coupled sewer rods of metal or wood used for clearing stoppages. Usually

available in 3–ft, 39–in, 4–ft, 5–ft and 6–ft lengths.

525

SUCTION HEAD : Condition that occurs when the liquid source is above the centerline of the pump.

SUCTION LIFT : Condition that occurs when the liquid source is below the centerline of the pump.

SUCTION STATIC HEAD : The difference in elevation between the liquid level of the source of supply

and the centerline of the pump. This head also includes any additional head that may be present at the

suction tank ﬂuid surface.

SUCTION STATIC LIFT : The same deﬁnition as the Suction Static head. This term is only used when

the pump centerline is above the suction tank ﬂuid surface.

SUPERNATANT : Liquid removed from settling sludge. Supernatant commonly refers to the liquid

between the sludge on the bottom and the scum on the surface of an anaerobic digester. The liquid is

usually returned to the inﬂuent wet well or to the primary clariﬁer.

SUPERVISORY CONTROL AND DATA ACQUISITION (SCADA) : A computer system for gathering

and analyzing real time data. SCADA systems are used to monitor and control a plant or equipment in

industry.

SURCHARGE : Sewers are surcharged when the supply of water to be carried is greater than the capacity

of the pipes to carry the ﬂow. The surface of the wastewater in manholes rises above the top of the sewer

pipe, and the sewer is under pressure or a head, rather than at atmospheric pressure.

SURCHARGED MANHOLE : A manhole in which the rate of the water entering is greater than the

capacity of the outlet under gravity ﬂow conditions. When the water in the manhole rises above the top of

the outlet pipe, the manhole is said to be “surcharged.”

SURFACE LOADING : Lagoon loading is rated organically in pounds of BOD per acre of surface area

per day. Northern climates require lower loading rates than warmer areas, because cold weather slows

down the stabilization processes of microorganisms. Treatment plants clariﬁers are rated hydraulically in

ﬂow (gpd) per surface area (sq ft).

SURFACTANT : Compounds that lower the surface tension (or interfacial tension) between two liquids

or between a liquid and a solid. Surfactants may act as detergents, wetting agents, emulsiﬁers, foaming

agents, and dispersants.

SUSPENDED SOLID : Solids that either ﬂoat on the surface or are suspended in water, wastewater, or

other liquids, and which are largely removable by laboratory ﬁltering.

SWAB : A circular sewer cleaning tool almost the same diameter as the pipe being cleaned. As a ﬁnal

cleaning procedure after a sewer line has been cleaned with a porcupine, a swab is pulled through the

sewer and the ﬂushing action of water ﬂowing around the tool cleans the line.

SWING CHECK VALVE : A check valve in which the closure element is a hinged clapper which swings

or rotates about a supporting shaft.

SYSTEM : Systems, as far as pumps are concerned, include all the piping with or without a pump, start-

ing at the inlet point (often the ﬂuid surface of the suction tank) and ending at the outlet point (often the

ﬂuid surface of the discharge tank).

SYSTEM CURVE : Is a plot of ﬂow vs. Total Head that satisﬁes the system requirements.

SYSTEM EQUATION : The equation for Total Head vs. ﬂow for a speciﬁc system.

SYSTEM REQUIREMENTS : Friction and system inlet and outlet conditions (i.e. velocity, elevation and

pressure).

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Chapter 25. Glossary

TAG LINE : A line, rope or cable that follows equipment through a sewer so that equipment can be pulled

back out if it encounters an obstruction or becomes stuck. Equipment is pulled forward with a pull line.

TAP : A small hole in a sewer where a wastewater service line from a building is connected (tapped) into a

lateral or branch sewer.

TELEVISION INSPECTION : An inspection of the inside of a sewer pipe made by pulling a closed–circuit

television camera through the pipe.

TERMINAL “LAMPHOLES” CLEANOUT : When a manhole is not provided at the upstream end of a

sewer main, a cleanout is usually provided. This is called a “terminal cleanout.”

TERTIARY WASTE TREATMENT : Following secondary treatment, the clariﬁed efﬂuent may require

additional aeration and/or other chemical treatment to destroy bacteria remaining from the secondary

treating stage, and to increase the content of dissolved oxygen needed for oxidation of the residual BOD.

Tertiary treatment can also be used to remove nitrogen and phosphorous.

THROTTLING : The intentional restriction of ﬂow by partially closing or opening a valve. A wide range

of throttling is accomplished automatically in regulators and control valves.

THRU: CONDUIT – An expression characterizing valves when in the open position, wherein the bore

presents a smooth uninterrupted interior surface across seat rings and thru the valve port, thus affording

minimum pressuredrop. There are no cavities or large gaps in the bore between seat rings and body clo-

sures or between seat rings and ball/gate. Consequently, there are no areas that can accumulate debris to

impede pipeline cleaning equipment or restrict the valve’s motion.

THRUST : The net force applied to a part in a particular direction – e.g., on the end of a valve stem.

TOP ENTRY : The design of a particular valve or regulator where the unit can be serviced or repaired by

leaving its body in the line, and its internals can be accessed by removing a top portion of the unit.

TORQUE : The turning effort required to operate a valve. Usually expressed in "pound–feet" and referred

to the stem nut, handwheel or operator pinion shaft.

TORQUE SWITCH : An electrical device on a motor operator which cuts off power to the operator when

allowable torque is exceeded, thus preventing damage to the valve and/or the operator.

TORSIONAL SPRING : A coiled spring which exerts a force by twisting about its axis rather than by

compression or elongation. The spring in a check valve slam retarder which is restrained at one end and

fastened tot he clappershaft on the other end. As the clapper opens, the spring resists the motion creating a

closing force. During a rapid decrease in ﬂow rate, the clapper is urged toward the closed position and is

virtually closed just prior to the instant of actual ﬂow reversal – thus slamming is avoided.

TOTAL DISSOLVED SOLIDS : Total dissolved solids are inorganic molecules of metals, minerals or

salts present in water at such a small size that you can’t see them. Because of their very small size, they

can be difﬁcult to remove with any technology other than ﬁne membrane ﬁltration technologies such as

Reverse Osmosis (RO).

TOTAL HEAD / TOTAL DYNAMIC HEAD : The amount of head produced by the pump. Calculated by

summing the static head, friction head, pressure head, and velocity head.

TOTAL KJELDAHL NITROGEN (TKN): Is a standard laboratory analysis which measures organically

bound nitrogen plus ammonia nitrogen.

TOTAL ORGANIC CARBON : A direct measurement of how much organic matter is in wastewater.

527

TOTAL SOLIDS : The total amount of solids in solution and suspension.

TOTAL STATIC HEAD : The difference between the discharge and suction static head including the

difference between the surface pressure of the discharge and suction tanks.

TOTAL SUSPENDED SOLIDS : Visible solids present in wastewater that can be ﬁltered out through

traditional physical treatment technologies. In the metal ﬁnishing industry, for example, FOG (fats, oils

and grease) and dirt particles might make up part of the total suspended solids.

TOTAL TOXIC ORGANICS : A wastewater parameter that refers to the entire amount of toxic organic

compounds present. EPA has developed a speciﬁc list of chemicals that are deﬁned as toxic organic

compounds.

TOXIC : A substance which is poisonous to a living organism.

TOXIC ORGANIC MANAGEMENT PLAN : A spill plan federally required for speciﬁc industries,

including metal ﬁnishing. It outlines what speciﬁc toxic organic compounds are used and how they are

disposed in a manner that prevents discharge to the sewer system.

TOXICITY : The relative degree of being poisonous or toxic. A condition which may exist in wastes and

will inhibit or destroy the growth or function of certain organisms.

TRANSPIRATION : See Evapotranspiration.

TRICKLING FILTER : An aerobic biological process used as secondary treatment of sewage. Efﬂuent

from the primary clariﬁer is distributed over a bed of rocks. As the liquid trickles over the rocks, a bio-

logical growth on the rocks breaks down the organic matter in the sewage. The efﬂuent is then taken to a

clariﬁer to remove biological matter coming from the ﬁlter.

TRIM : Commonly refers to the valve’s working parts and to their materials. Usually includes seat ring

sealing surfaces, closure element sealing surfaces, stems, and back seats. Trim numbers which specifythe

materials are deﬁned in API 600 and API 602.

TRIPLE ECCENTRIC (BUTTERFLY VALVES) : A particular design of a butterﬂy valve where the stem

is located behind the disc, below the centerline of the disc, and its cone axis is offset from the centerline of

the disc. This particular designis capable of a very tight shutoff at temperatures well above 100°F.

TRUNNION : That part of a ball valve which holds the ball on a ﬁxed vertical axis and about which the

ball turns. The torque requirement of a trunnion mounted ball valve is signiﬁcantly less than that for a

ﬂoatingball design.

TSS : Abbreviation for TOTAL SUSPENDED SOLIDS, a test measuring the amount of ﬁlterable solids in

wastewater.

TURBID : Having a cloudy or muddy appearance.

TURBIDITY : The cloudy appearance of water caused by the presence of suspended and colloidal matter.

A turbidity measurement is used to indicate the clarity of water.

TURNS TO OPERATE : The number of complete revolutions of a handwheel or the pinion shaft of a gear

operator required to stroke a valve from fully open to fully closed or vice versa.

TWO: WAY SPHERE–LOK – A Sphere–Lok with two ports.

U: CUP (RING–PACKING) – A "U" cross–section ring located on the tail end of certain ball valve seats

to retain the grease in an emergency seat seal system.

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Chapter 25. Glossary

ULTRA FILTRATION : Type of membrane ﬁltration that separates suspended solids and solutes of high

molecular from a liquid and low molecular weight solutes. The ultra ﬁltration separation process is used

in industry processes, water treatment and research for purifying and concentrating macromolecular

solutions.

ULTRAVIOLET : In some industries, ultraviolet light is used to sterilize water treated wastewater prior to

reuse or recycling. UV light keeps algae and other bacteria from growing in the recycled wastewater.

UNION BONNET : A type of valve construction in which the bonnet is held on by a union nut with

threads on the body.

UV : Ultraviolet light. UV is useful as a method of disinfection. It leaves no residual and is often used

where no chlorine residual (or a very low residual) is allowed to be discharged.

UV DISINFECTION (UVGI) : Ultraviolet germicidal irradiation (UVGI) is a disinfection method that

uses ultraviolet (UV) light at sufﬁciently short wavelength to kill microorganisms. It is used in a variety

of applications, such as food, air and water puriﬁcation. UVGI utilizes short–wavelength ultraviolet

radiation (UV–C) that is harmful to microorganisms. It is effective in destroying the nucleic acids in

these organisms so that their DNA is disrupted by the UV radiation, leaving them unable to perform vital

cellular functions.

VALVE : A device used to control the ﬂow of ﬂuid contained in a pipe line.

VAPOR PRESSURE : The pressure at which a liquid boils at a speciﬁed temperature.

VARIABLE FREQUENCY DRIVE (VFD) : A system for controlling the rotational speed of an alter-

nating current (AC) electric motor by controlling the frequency of the electrical power supplied to the

motor.

VARIABLE FREQUENCY DRIVE : A variable–frequency drive (VFD) is a system for controlling the

rotational speed of an alternating current (AC) electric motor by controlling the frequency of the electrical

power supplied to the motor. A variable frequency drive is a speciﬁc type of adjustable–speed drive.

Variable–frequency drives are also known as adjustable–frequency drives (AFD), variable–speed drives

(VSD), AC drives, micro drives or inverter drives. Since the voltage is varied along with frequency, these

are sometimes also called VVVF (variable voltage variable frequency) drives. Variable–frequency drives

are widely used. For example, in water booster stations, pump speed is controlled by the VFD based on

system demand.

VARIABLE ORIFICE : A small variable proﬁle valve put in a ﬂow line and used with a pilot to restrict

the ﬂow into the pilot and make the pilot more or less sensitive to changing conditions.

VDS : VALVE DATA SHEET – A data sheet deﬁning the minimum level of a valve design, including the

materials, testing, inspection, and certiﬁcation requirements.

VELOCITY HEAD DIFFERENCE : The difference in velocity head between the outlet and inlet of the

system.

VENT PLUG : (VENT PLUG ASSEMBLY) – (SAFETY VENT PLUG) – A special pipe plug having a

small allen–wrench operated vent valve. These special plugs are located at the bottom of most ball valves.

With the line valve closed (and under pressure) the body cavity pressurecan be vented thru this small valve

to check tightness of seat seals or to make minor repairs. Having vented the body pressure, the vent plug

may be removed to blow out debris and foreign material or to ﬂush the body cavity. On some gate valves,

the vent plug is installed on the bonnet for the sole purpose of venting the body. Such valves have separate

529

drain valves.

VENTURI VALVE : A reduced bore valve. A valve having a bore smaller in diameter than the inlet or

outlet. For example, an 8"x 6" x 8" ball valve has 8" inlet and outlet connections while the ball and seats

are 6". The ﬂow through a venture valve will be reduced because of the smaller port. Venturi valves can

often be economically substituted for plug valves.

VERTICAL TURBINE PUMP : A vertical turbine pump is a centrifugal type pump, often with multiple

stages, where the motor is set at ground level and connect via shaft to the pump below. Used as well

pumps for irrigation, they can also pump from rivers, lakes and other bodies of water.

VISCOSITY : A property, which measures a ﬂuid’s resistance to movement. The resistance is caused

by friction between the ﬂuid and the boundary wall and internally by the ﬂuid layers moving at different

velocities.

VOLATILE ORGANIC COMPOUNDS : In wastewater, VOCs typically show up as cleaning solvents.

These chemicals can kill the microbes in POTWs (publicly owned treatment work) if they are discharged

in large quantities, so they are carefully regulated. They’re challenging to treat because they dissolve in

water.

VOLATILE SOLIDS : Those solids in water, wastewater, or other liquids that are lost on ignition of the

dry solids at 550oC.

VOLATILE SOLIDS : Those solids in water, wastewater, or other liquids that are lost on ignition of the

dry solids at 550oC.

W.O. : WRENCH OPERATED – The operation of a valve by means of a handle or lever. Used on smaller

size and lower pressure class valves.

WALL THICKNES : The thickness of the wall of the pressure vessel or valve. For steel valves, minimum

thickness requirements are deﬁned in ASME B16.34, API 600, and API 602.

WASTE ACTIVATED SLUDGE : That portion of sludge from the secondary clariﬁer in the activated

sludge process that is wasted to avoid a buildup of solids in the system.

WASTE TREATMENT SLUDGE : A series of tanks, screens, ﬁlters, and other processes by which most

pollutants are removed from water.

WASTEWATER : The used water and solids from a community that ﬂow to a treatment plant. Storm

water, surface water, and groundwater inﬁltration also may be included in the wastewater that enters a

wastewater treatment plant. The term “sewage” usually refers to household wastes, but this word is being

replaced by the term “wastewater.

WASTEWATER TREATMENT PLANT : A city, municipal or industrial facility that’s treating wastewa-

ter.

WATER HAMMER : The physical effect, often accompanied by loud banging, produced by pressure

waves generated within the piping by rapid change of velocity in a liquid system.

WATER POLLUTION : A general term signifying the introduction into water of micro–organisms, chemi-

cals, wastes, or sewage which renders the water unﬁt for it’s intended use.

WATER QUALITY ACT : Montana’s primary water pollution control legislation that parallels the federal

CLEAN WATER ACT. It establishes the public policy for Montana to – 1) conserve water resources by

protecting, maintaining and improving water quality for all its beneﬁcial uses, and 2) provides a compre-

530

Chapter 25. Glossary

hensive program for the prevention, abatement and control of water pollution.

WATER TREATMENT FACILITY : A city or municipal water treatment facility that’s treating water you

drink or use in an industrial process.

WEDGE GATE : A gate whose seating surfaces are inclined to the direction of closing thrust so that

mechanical force on the stem produces tight contact with the inclined seat rings.

WEIR : (1) A wall or plate placed in an open channel and used to measure the ﬂow of water. The depth

of the ﬂow over the weir can be used to calculate the ﬂow rate, or a chart or conversion table may be used.

(2) A wall or obstruction used to control ﬂow (from settling tanks and clariﬁers) to assure a uniform ﬂow

rate and avoid short–circuiting.

WELL PUMP : Well pumps are a centrifugal, submersible type of pump used for bringing underground

water up to the surface for domestic use. They can consist of one or several “stages” depending on the

well depth and desired discharge pressure. Electrically powered the motor is typically on the bottom of

the pump with the suction in the middle and the water is pumped upwards through the impeller(s) and

upwards toward the surface.

WET OXIDATION : A method of treating or conditioning sludge before the water is removed. Com-

pressed air is blown into the sludge; the air and sludge mixture is fed into a pressure vessel where the

organic material is stabilized.

WET WELL : A compartment or tank in which wastewater is collected. The suction pipe of a pump may

be connected to the wet well or a submersible pump may be located in the wet well.

WETTED PARTS : A term used for any part that comes into contact with the process ﬂuid. These parts

must be checked for chemical compatibility with the process ﬂuid.

WORK : The energy required to drive the ﬂuid through the system. A measure of a liquid’s resistance to

ﬂow. Essentially it’s a how thick the liquid is. The viscosity determines the type of pump used, the speed

it can run at, and with gear pumps, the internal clearances required.

YOKE : That part of a gate valve which serves as a spacer between the bonnet and the operator or actua-

tor.

ZINC : A heavy metal commonly regulated by wastewater permits. It is widely present in all industries

and can be difﬁcult to treat to low levels through typical physical/chemical treatment technologies. It

is very important to determine the sources of zinc in process wastewater in order to adequately control

discharge levels. Other heavy metals include – Arsenic (As), Cadmium (Cd), Chromium(Cr), Copper

(Cu), Lead (Pb), and Nickel (Ni).

ZOOGLEAL MASS : Jelly like masses of bacteria found in both the trickling ﬁlter and activated sludge

processes. See also Biomass.